## Normal Distribution Problems with Solutions

Problems and applications on normal distributions are presented. The solutions to these problems are at the bottom of the page. An online normal probability calculator and an inverse normal probability calculator may be useful to check your answers.

## Problems with Solutions

• X is a normally distributed variable with mean = 30 and standard deviation = 4. Find the probabilities a) P(X < 40) b) P(X > 21) c) P(30 < X < 35)
• A radar unit is used to measure the speeds of cars on a motorway. The speeds are normally distributed with a mean of 90 km/hr and a standard deviation of 10 km/hr. What is the probability that a car picked at random is traveling at more than 100 km/hr?
• For certain types of computers, the length of time between charges of the battery is normally distributed with a mean of 50 hours and a standard deviation of 15 hours. John owns one of these computers and wants to know the probability that the length of time will be between 50 and 70 hours.
• Entry to a certain University is determined by a national test. The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100. Tom wants to be admitted to this university and he knows that he must score better than at least 70% of the students who took the test. Tom takes the test and scores 585. Will he be admitted to this university?
• The length of similar components produced by a company is approximated by a normal distribution model with a mean of 5 cm and a standard deviation of 0.02 cm. If a component is chosen at random a) what is the probability that the length of this component is between 4.98 and 5.02 cm? b) What is the probability that the length of this component is between 4.96 and 5.04 cm?
• The length of life of an instrument produced by a machine has a normal distribution with a mean of 12 months and a standard deviation of 2 months. Find the probability that an instrument produced by this machine will last a) less than 7 months. b) between 7 and 12 months.
• The time taken to assemble a car in a certain plant is a random variable having a normal distribution of 20 hours and a standard deviation of 2 hours. What is the probability that a car can be assembled at this plant in a period of time a) less than 19.5 hours? b) between 20 and 22 hours?
• A large group of students took a test in Physics and the final grades have a mean of 70 and a standard deviation of 10. If we can approximate the distribution of these grades by a normal distribution, what percent of the students a) scored higher than 80? b) Should pass the test (grades60)? c) Should fail the test (grades<60)?
• The annual salaries of employees in a large company are approximately normally distributed with a mean of $50,000 and a standard deviation of$20,000. a) What percent of people earn less than $40,000? b) What percent of people earn between$45,000 and $65,000? c) What percent of people earn more than$70,000?

## Solutions to the Above Problems

• Note: What is meant here by area is the area under the standard normal curve. a) For x = 40, the z-value z = (40 - 30) / 4 = 2.5 Hence P(x < 40) = P(z < 2.5) = [area to the left of 2.5] = 0.9938 b) For x = 21, z = (21 - 30) / 4 = -2.25 Hence P(x > 21) = P(z > -2.25) = [total area] - [area to the left of -2.25] = 1 - 0.0122 = 0.9878 c) For x = 30 , z = (30 - 30) / 4 = 0 and for x = 35, z = (35 - 30) / 4 = 1.25 Hence P(30 < x < 35) = P(0 < z < 1.25) = [area to the left of z = 1.25] - [area to the left of 0] = 0.8944 - 0.5 = 0.3944
• Let x be the random variable that represents the speed of cars. x has = 90 and = 10. We have to find the probability that x is higher than 100 or P(x > 100) For x = 100 , z = (100 - 90) / 10 = 1 P(x > 90) = P(z > 1) = [total area] - [area to the left of z = 1] = 1 - 0.8413 = 0.1587 The probability that a car selected at random has a speed greater than 100 km/hr is equal to 0.1587
• Let x be the random variable that represents the length of time. It has a mean of 50 and a standard deviation of 15. We have to find the probability that x is between 50 and 70 or P( 50< x < 70) For x = 50 , z = (50 - 50) / 15 = 0 For x = 70 , z = (70 - 50) / 15 = 1.33 (rounded to 2 decimal places) P( 50< x < 70) = P( 0< z < 1.33) = [area to the left of z = 1.33] - [area to the left of z = 0] = 0.9082 - 0.5 = 0.4082 The probability that John's computer has a length of time between 50 and 70 hours is equal to 0.4082.
• Let x be the random variable that represents the scores. x is normally distributed with a mean of 500 and a standard deviation of 100. The total area under the normal curve represents the total number of students who took the test. If we multiply the values of the areas under the curve by 100, we obtain percentages. For x = 585 , z = (585 - 500) / 100 = 0.85 The proportion P of students who scored below 585 is given by P = [area to the left of z = 0.85] = 0.8023 = 80.23% Tom scored better than 80.23% of the students who took the test and he will be admitted to this University.
• a) P(4.98 < x < 5.02) = P(-1 < z < 1) = 0.6826 b) P(4.96 < x < 5.04) = P(-2 < z < 2) = 0.9544
• a) P(x < 7) = P(z < -2.5) = 0.0062 b) P(7 < x < 12) = P(-2.5 < z < 0) = 0.4938
• a) P(x < 19.5) = P(z < -0.25) = 0.4013 b) P(20 < x < 22) = P(0 < z < 1) = 0.3413
• a) For x = 80, z = 1 Area to the right (higher than) z = 1 is equal to 0.1586 = 15.87% scored more that 80. b) For x = 60, z = -1 The area to the right of z = -1 is equal to 0.8413 = 84.13% should pass the test. c) 100% - 84.13% = 15.87% should fail the test.
• a) For x = 40000, z = -0.5 The area to the left (less than) of z = -0.5 is equal to 0.3085 = 30.85% earn less than $40,000. b) For x = 45000 , z = -0.25 and for x = 65000, z = 0.75 The area between z = -0.25 and z = 0.75 is equal to 0.3720 = 37.20 earning between$45,000 and $65,000. c) For x = 70000, z = 1 The area to the right (higher) of z = 1 is equal to 0.1586 = 15.86% earning more than$70,000.

• Normal Distribution Definition
• Elementary statistics and probabilities

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• The Standard Normal Distribution | Calculator, Examples & Uses

## The Standard Normal Distribution | Calculator, Examples & Uses

Published on November 5, 2020 by Pritha Bhandari . Revised on June 21, 2023.

The standard normal distribution , also called the z -distribution , is a special normal distribution where the mean is 0 and the standard deviation is 1.

Any normal distribution can be standardized by converting its values into z scores. Z scores tell you how many standard deviations from the mean each value lies.

Converting a normal distribution into a z -distribution allows you to calculate the probability of certain values occurring and to compare different data sets.

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All normal distributions , like the standard normal distribution, are unimodal and symmetrically distributed with a bell-shaped curve. However, a normal distribution can take on any value as its mean and standard deviation. In the standard normal distribution, the mean and standard deviation are always fixed.

Every normal distribution is a version of the standard normal distribution that’s been stretched or squeezed and moved horizontally right or left.

The mean determines where the curve is centered. Increasing the mean moves the curve right, while decreasing it moves the curve left.

The standard deviation stretches or squeezes the curve. A small standard deviation results in a narrow curve, while a large standard deviation leads to a wide curve.

When you standardize a normal distribution, the mean becomes 0 and the standard deviation becomes 1. This allows you to easily calculate the probability of certain values occurring in your distribution, or to compare data sets with different means and standard deviations.

While data points are referred to as x in a normal distribution, they are called z or z scores in the z distribution. A z score is a standard score that tells you how many standard deviations away from the mean an individual value ( x ) lies:

• A positive z score means that your x value is greater than the mean.
• A negative z score means that your x value is less than the mean.
• A z score of zero means that your x value is equal to the mean.

Converting a normal distribution into the standard normal distribution allows you to:

• Compare scores on different distributions with different means and standard deviations.
• Normalize scores for statistical decision-making (e.g., grading on a curve).
• Find the probability of observations in a distribution falling above or below a given value.
• Find the probability that a sample mean significantly differs from a known population mean.

## How to calculate a z score

To standardize a value from a normal distribution, convert the individual value into a z -score:

• Subtract the mean from your individual value.
• Divide the difference by the standard deviation.

To standardize your data, you first find the z score for 1380. The z score tells you how many standard deviations away 1380 is from the mean.

The  z score for a value of 1380 is 1.53 . That means 1380 is 1.53 standard deviations from the mean of your distribution.

The standard normal distribution is a probability distribution , so the area under the curve between two points tells you the probability of variables taking on a range of values. The total area under the curve is 1 or 100%.

Every z score has an associated p value that tells you the probability of all values below or above that z score occuring. This is the area under the curve left or right of that z score.

## Z tests and p values

The z score is the test statistic used in a z test . The z test is used to compare the means of two groups, or to compare the mean of a group to a set value. Its null hypothesis typically assumes no difference between groups.

The area under the curve to the right of a z score is the p value, and it’s the likelihood of your observation occurring if the null hypothesis is true.

Usually, a p value of 0.05 or less means that your results are unlikely to have arisen by chance; it indicates a statistically significant effect.

By converting a value in a normal distribution into a z score, you can easily find the p value for a z test.

How to use a z table

Once you have a z score, you can look up the corresponding probability in a z table .

In a z table, the area under the curve is reported for every z value between -4 and 4 at intervals of 0.01.

There are a few different formats for the z table. Here, we use a portion of the cumulative table. This table tells you the total area under the curve up to a given z score—this area is equal to the probability of values below that z score occurring.

The first column of a z table contains the z score up to the first decimal place. The top row of the table gives the second decimal place.

To find the corresponding area under the curve (probability) for a z score:

• Go down to the row with the first two digits of your z score.
• Go across to the column with the same third digit as your z  score.
• Find the value at the intersection of the row and column from the previous steps.

To find the shaded area, you take away 0.937 from 1, which is the total area under the curve.

Probability of x > 1380 = 1 − 0.937 = 0.063

Let’s walk through an invented research example to better understand how the standard normal distribution works.

As a sleep researcher, you’re curious about how sleep habits changed during COVID-19 lockdowns. You collect sleep duration data from a sample during a full lockdown.

Before the lockdown, the population mean was 6.5 hours of sleep. The lockdown sample mean is 7.62.

To assess whether your sample mean significantly differs from the pre-lockdown population mean, you perform a z test :

• First, you calculate a z score for the sample mean value.
• Then, you find the p value for your z score using a z table.

## Step 1: Calculate a z -score

To compare sleep duration during and before the lockdown, you convert your lockdown sample mean into a z score using the pre-lockdown population mean and standard deviation.

A z score of 2.24 means that your sample mean is 2.24 standard deviations greater than the population mean.

## Step 2: Find the  p value

To find the probability of your sample mean z score of 2.24 or less occurring, you use the  z table to find the value at the intersection of row 2.2 and column +0.04.

The table tells you that the area under the curve up to or below your z score is 0.9874. This means that your sample’s mean sleep duration is higher than about 98.74% of the population’s mean sleep duration pre-lockdown.

To find the p value to assess whether the sample differs from the population, you calculate the area under the curve above or to the right of your z score. Since the total area under the curve is 1, you subtract the area under the curve below your z score from 1.

A p value of less than 0.05 or 5% means that the sample significantly differs from the population.

Probability of z > 2.24 = 1 − 0.9874 = 0.0126 or 1.26%

With a p value of less than 0.05, you can conclude that average sleep duration in the COVID-19 lockdown was significantly higher than the pre-lockdown average.

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

• Student’s t table
• Student’s t distribution
• Descriptive statistics
• Measures of central tendency
• Correlation coefficient

Methodology

• Cluster sampling
• Stratified sampling
• Types of interviews
• Cohort study
• Thematic analysis

Research bias

• Implicit bias
• Cognitive bias
• Survivorship bias
• Availability heuristic
• Nonresponse bias
• Regression to the mean

In a normal distribution , data are symmetrically distributed with no skew. Most values cluster around a central region, with values tapering off as they go further away from the center.

The measures of central tendency (mean, mode, and median) are exactly the same in a normal distribution.

The standard normal distribution , also called the z -distribution, is a special normal distribution where the mean is 0 and the standard deviation is 1.

Any normal distribution can be converted into the standard normal distribution by turning the individual values into z -scores. In a z -distribution, z -scores tell you how many standard deviations away from the mean each value lies.

The empirical rule, or the 68-95-99.7 rule, tells you where most of the values lie in a normal distribution :

• Around 68% of values are within 1 standard deviation of the mean.
• Around 95% of values are within 2 standard deviations of the mean.
• Around 99.7% of values are within 3 standard deviations of the mean.

The empirical rule is a quick way to get an overview of your data and check for any outliers or extreme values that don’t follow this pattern.

The t -distribution gives more probability to observations in the tails of the distribution than the standard normal distribution (a.k.a. the z -distribution).

In this way, the t -distribution is more conservative than the standard normal distribution: to reach the same level of confidence or statistical significance , you will need to include a wider range of the data.

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## Normal Distribution

In these lessons, we learn the characteristics of the normal distribution and its applications.

Related Pages Normal Distribution Normal Distribution: Probability Standard Deviation More Lessons for Statistics Math Worksheets

What is the Normal Distribution? Probably the most widely known and used of all distributions is the normal distribution. It fits many human characteristics, such as height, weight, speed etc. Many living things in nature, such as trees, animals and insects have many characteristics that are normally distributed. Many variables in business and industry are also normally distributed.

Discovery of the normal curve is generally credited to Karl Gauss (1777 – 1855), who recognized that the errors of repeated measurement of objects are often normally distributed. Sometimes, the normal distribution is also called the Gaussian distribution.

The normal distribution has the following characteristics:

• It is a continuous distribution
• It is symmetrical about the mean. Each half of the distribution is a mirror image of the other half.
• It is asymptotic to the horizontal axis. That is, it does not touch the x -axis and it goes on forever in each direction.
• It is unimodal. The normal curve is sometimes called a bell-shaped curve. All the values are “bunched up” in only one portion of the graph – the center of the curve.
• It is a family of curves. Every unique value of the mean and every unique value of the standard deviation result in a different normal curve.
• The area under the curve is 1. The area under the curve yields the probabilities, so the total of all probabilities for a normal distribution is 1. Since the distribution is symmetric, the area of the distribution on each side of the mean is 0.5.

What is the Probability density function of the normal distribution? The normal distribution is described by two parameters: the mean, μ, and the standard deviation, σ. We write X - N(μ, σ 2 ).

The following diagram shows the formula for Normal Distribution. Scroll down the page for more examples and solutions on using the normal distribution formula.

Since the formula is so complex, using it to determine area under the curve is cumbersome and time consuming. Instead, tables and software are used to find the probabilities for the normal distribution. This will be discussed in the lesson on Z-Score .

Normal Distribution The 68-95-99.7 Rule

• Approximately 68% of the data falls ±1 standard deviation from the mean.
• Approximately 95% of the data falls ±2 standard deviation from the mean.
• Approximately 99.7% of the data falls ±3 standard deviation from the mean.

In a call center, the distribution of the number of phone calls answered each day by each of the 12 receptionists is bell-shaped and has a mean of 63 and a standard deviation of 3. Use the empirical rule, what is the approximate percentage of daily phone calls numbering between 60 and 66?

The scores of a midterm are normally distributed with a mean of 85% and a standard deviation of 6%. Find the percentage of the class that score above and below the given score. Use the 68-95-99.7 rule from the text. a) Score: 91% b) Score: 73%

The following video explores the normal distribution

Presentation on spreadsheet to show that the normal distribution approximates the binomial distribution for a large number of trials.

Teach yourself statistics

## The Normal Distribution

In this lesson, we describe the normal distribution, a continuous probability distribution that is widely used in statistics to compute probabilities associated with various naturally-occurring events.

## The Normal Equation

The normal distribution is defined by the following equation:

The Normal Equation . The value of the random variable Y is:

Y = { 1/[ σ * sqrt(2π) ] } * e -(X - μ) 2 /2σ 2

where X is a normal random variable, μ is the mean, σ is the standard deviation, π is approximately 3.14159, and e is approximately 2.71828.

The random variable X in the normal equation is called the normal random variable . It can range in value from minus to plus infinity. The normal equation is the probability density function for the normal distribution.

## The Normal Curve

When X and Y values from the normal equation are graphed on an X-Y scatter chart, all normal distributions look like a symmetric, bell-shaped curve, as shown below.

Smaller standard deviation

Bigger standard deviation

The location and shape of the curve for the normal distribution depends on two factors - the mean and the standard deviation.

• The center of the curve is located on the X-axis at the mean of the distribution.
• The standard deviation determines the shape (height and width) of the curve. When the standard deviation is small, the curve is tall and narrow; and when the standard deviation is big, the curve is short and wide (see above)

## Probability and the Normal Curve

The normal distribution is a continuous probability distribution. This has several implications for probability.

• The total area under the normal curve is equal to 1. And, the probability that a normal random variable X is less than positive infinity is equal to 1; that is, P(X < ∞) = 1.
• The probability that a normal random variable X equals any particular value is 0. Here's why. The normal random variable X can take any value between minus and plus infinity - an infinite number of values. The probability of selecting any single value from an infinitely large set of values is always zero.
• Let a equal any real number. The probability that X is greater than a equals the area under the normal curve bounded by a and plus infinity (as indicated by the non-shaded area in the figure below).
• The probability that X is less than a equals the area under the normal curve bounded by a and minus infinity (as indicated by the shaded area in the figure below).

Additionally, every normal curve (regardless of its mean or standard deviation) conforms to the following "rule".

• About 68% of the area under the curve falls within 1 standard deviation of the mean.
• About 95% of the area under the curve falls within 2 standard deviations of the mean.
• About 99.7% of the area under the curve falls within 3 standard deviations of the mean.

Collectively, these points are known as the empirical rule or the 68-95-99.7 rule . Clearly, given a normal distribution, most outcomes will be within 3 standard deviations of the mean.

## Why the Normal Curve is Useful

The values observed for some natural phenomena (height, weight, IQ, blood pressure, etc.) follow an approximate normal distribution. For those phenomena, the normal distribution provides a useful frame of reference for computing probability.

To illustrate how the normal distribution provides a useful frame of reference for probability, consider this example. Suppose we weighed all of the brown mushrooms harvested on a farm in a single planting season. We might find that the mean weight of a mushroom was 60 grams, and the standard deviation was 4 grams. We could plot mushroom weight on a histogram, like so:

Notice that this histogram is symmetric with a single peak in the center, not too different from the bell-shaped curve of a normal distribution. If we display the histogram above a normal curve having a mean of 6 and standard deviation of 4, it is easy to see the resemblance.

Given the similarities, you might suspect that the normal distribution could be useful in predicting probabilities involving mushroom weight. And you would be right! (For an example that shows how to predict probabilities associated with mushrooom weight, see Problem 3 below.)

Bottom line: Suppose X is a random variable that is distributed roughly normally in the population. If you know the mean and standard deviation of X, you can compute a cumulative probability for X. Specifically, you can compute P(X < x) and P(X > x).

## How to Find Probability

To find a cumulative probability for a normal random variable, world-class statisticians can use the normal equation described earlier (plus a little calculus). However, the rest of us use one of the following:

• A graphing calculator.
• An online probability calculator, such as Stat Trek's Normal Distribution Calculator .
• A normal distribution probability table (found in the appendix of most introductory statistics texts).

In the examples below, we use Stat Trek's Normal Distribution Calculator to calculate probability. In the next lesson, we use normal distribution probability tables.

## Normal Distribution Calculator

The normal distribution calculator solves common statistical problems, based on the normal distribution. The calculator computes cumulative probabilities, based on three simple inputs. Simple instructions guide you to an accurate solution, quickly and easily. If anything is unclear, frequently-asked questions and sample problems provide straightforward explanations. The calculator is free. It can found in the Stat Trek main menu under the Stat Tools tab. Or you can tap the button below.

Problem 1 An average light bulb manufactured by the Acme Corporation lasts 300 days with a standard deviation of 50 days. Assuming that bulb life is normally distributed, what is the probability that an Acme light bulb will last at most 365 days?

Solution: Given a mean score of 300 days and a standard deviation of 50 days, we want to find the cumulative probability that bulb life is less than or equal to 365 days. Thus, we know the following:

• The value of the normal random variable is 365 days.
• The mean is equal to 300 days.
• The standard deviation is equal to 50 days.

We enter these values into the Normal Distribution Calculator and compute the cumulative probability.

The answer is: P( X < 365) = 0.90319. Hence, there is about a 90% chance that a light bulb will burn out within 365 days.

Suppose scores on an IQ test are normally distributed. If the test has a mean of 100 and a standard deviation of 10, what is the probability that a person who takes the test will score between 90 and 110?

Solution: Here, we want to know the probability that the test score falls between 90 and 110. The "trick" to solving this problem is to realize the following:

P( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 )

We use the Normal Distribution Calculator to compute both probabilities on the right side of the above equation.

• To compute P( X < 110 ), we enter the following inputs into the calculator: The raw score value of the normal random variable is 110, the mean is 100, and the standard deviation is 10. We find that P( X < 110 ) is about 0.84.
• To compute P( X < 90 ), we enter the following inputs into the calculator: The raw score value of the normal random variable is 90, the mean is 100, and the standard deviation is 10. We find that P( X < 90 ) is about 0.16.

We use these findings to compute our final answer as follows:

P( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 ) P( 90 < X < 110 ) = 0.84 - 0.16 P( 90 < X < 110 ) = 0.68

Thus, about 68% of the test scores will fall between 90 and 110, as predicted by the 68-95-99.7 rule .

Suppose a farmer collects a random sample of fully-developed mushrooms. He finds that the mean weight of a mushroom in his sample is 60 grams, and the standard deviation is 4 grams. Suppose further that his buyer will only purchase mushrooms bigger than 57 grams.

What is the probability that a mushroom harvested by this farmer will be smaller than 57 grams?

Solution: Given a mean weight of 60 grams and a standard deviation of 4 grams, we want to find the cumulative probability that a mushroom will weigh less than or equal to 57 grams. Thus, we know the following:

• The raw mushroom weight of interest is 57 grams.
• The mean mushroom weight is 60 grams.
• The standard deviation is 4 grams.

The answer is: P( X ≤ 57) = 0.22663. Hence, there is about a 23% chance that a mushroom in this farmer's crop will weigh less than 57 grams.

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## 6.2: Applications of the Normal Distribution

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The shaded area in the following graph indicates the area to the left of $$x$$. This area is represented by the probability $$P(X < x)$$. Normal tables, computers, and calculators provide or calculate the probability $$P(X < x)$$.

The area to the right is then $$P(X > x) = 1 – P(X < x)$$. Remember, $$P(X < x) =$$ Area to the left of the vertical line through $$x$$. $$P(X > x) = 1 – P(X < x) =$$ Area to the right of the vertical line through $$x$$. $$P(X < x)$$ is the same as $$P(X \leq x)$$ and $$P(X > x)$$ is the same as $$P(X \geq x)$$ for continuous distributions.

## Calculations of Probabilities

Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators.To calculate the probability, use the probability tables provided in [link] without the use of technology. The tables include instructions for how to use them.

## Example $$\PageIndex{1}$$

If the area to the left is 0.0228, then the area to the right is $$1 - 0.0228 = 0.9772$$.

## Exercise $$\PageIndex{1}$$

If the area to the left of $$x$$ is $$0.012$$, then what is the area to the right?

$$1 - 0.012 = 0.988$$

## Example $$\PageIndex{2}$$

The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.

• Find the probability that a randomly selected student scored more than 65 on the exam.
• Find the probability that a randomly selected student scored less than 85.
• Find the 90 th percentile (that is, find the score $$k$$ that has 90% of the scores below k and 10% of the scores above $$k$$).
• Find the 70 th percentile (that is, find the score $$k$$ such that 70% of scores are below $$k$$ and 30% of the scores are above $$k$$).

a. Let $$X$$ = a score on the final exam. $$X \sim N(63, 5)$$, where $$\mu = 63$$ and $$\sigma = 5$$

Draw a graph.

Then, find $$P(x > 65)$$.

$P(x > 65) = 0.3446\nonumber$

The probability that any student selected at random scores more than 65 is 0.3446.

## USING THE TI-83, 83+, 84, 84+ CALCULATOR

Go into 2nd DISTR .

After pressing 2nd DISTR , press 2:normalcdf .

The syntax for the instructions are as follows:

normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 10 99 ) by pressing 1 , the EE key (a 2nd key) and then 99 . Or, you can enter 10^ 99 instead. The number 10 99 is way out in the right tail of the normal curve. We are calculating the area between 65 and 10 99 . In some instances, the lower number of the area might be –1E99 (= –10 99 ). The number –10 99 is way out in the left tail of the normal curve.

## Historical Note

The TI probability program calculates a $$z$$-score and then the probability from the $$z$$-score. Before technology, the $$z$$-score was looked up in a standard normal probability table (because the math involved is too cumbersome) to find the probability. In this example, a standard normal table with area to the left of the $$z$$-score was used. You calculate the $$z$$-score and look up the area to the left. The probability is the area to the right.

$z = 65 – 63565 – 635 = 0.4\nonumber$

Area to the left is 0.6554.

$P(x > 65) = P(z > 0.4) = 1 – 0.6554 = 0.3446\nonumber$

Find the percentile for a student scoring 65:

*Press 2nd Distr *Press 2:normalcdf ( *Enter lower bound, upper bound, mean, standard deviation followed by ) *Press ENTER . For this Example, the steps are 2nd Distr 2:normalcdf (65,1,2nd EE,99,63,5) ENTER The probability that a selected student scored more than 65 is 0.3446. To find the probability that a selected student scored more than 65, subtract the percentile from 1.

b. Draw a graph.

Then find $$P(x < 85)$$, and shade the graph.

Using a computer or calculator, find $$P(x < 85) = 1$$.

$$\text{normalcdf}(0,85,63,5) = 1$$ (rounds to one)

The probability that one student scores less than 85 is approximately one (or 100%).

c. Find the 90 th percentile. For each problem or part of a problem, draw a new graph. Draw the $$x$$-axis. Shade the area that corresponds to the 90 th percentile.

Let $$k =$$ the 90 th percentile. The variable $$k$$ is located on the $$x$$-axis. $$P(x < k)$$ is the area to the left of $$k$$. The 90 th percentile $$k$$ separates the exam scores into those that are the same or lower than $$k$$ and those that are the same or higher. Ninety percent of the test scores are the same or lower than $$k$$, and ten percent are the same or higher. The variable $$k$$ is often called a critical value.

$$k = 69.4$$

The 90 th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above. To get this answer on the calculator, follow this step:

invNorm in 2nd DISTR . invNorm(area to the left, mean, standard deviation)

For this problem, $$\text{invNorm}(0.90,63,5) = 69.4$$

d. Find the 70 th percentile.

Draw a new graph and label it appropriately. $$k = 65.6$$

The 70 th percentile is 65.6. This means that 70% of the test scores fall at or below 65.6 and 30% fall at or above.

$$\text{invNorm}(0.70,63,5) = 65.6$$

## Exercise $$\PageIndex{2}$$

The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a randomly selected golfer scored less than 65.

$$\text{normalcdf}(10^{99},65,68,3) = 0.1587$$

## Example $$\PageIndex{3}$$

A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour.

• Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.
• Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.

a. Let $$X =$$ the amount of time (in hours) a household personal computer is used for entertainment. $$X \sim N(2, 0.5)$$ where $$\mu = 2$$ and $$\sigma = 0.5$$.

Find $$P(1.8 < x < 2.75)$$.

The probability for which you are looking is the area between $$x = 1.8$$ and $$x = 2.75$$. $$P(1.8 < x < 2.75) = 0.5886$$

$\text{normalcdf}(1.8,2.75,2,0.5) = 0.5886\nonumber$

The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.

To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25 th percentile, $$k$$, where $$P(x < k) = 0.25$$.

$\text{invNorm}(0.25,2,0.5) = 1.66\nonumber$

The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.

## Exercise $$\PageIndex{3}$$

The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70.

$$\text{normalcdf}(66,70,68,3) = 0.4950$$

## Example $$\PageIndex{4}$$

There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.

• Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.
• Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.
• Find the 80 th percentile of this distribution, and interpret it in a complete sentence.
• $$\text{normalcdf}(23,64.7,36.9,13.9) = 0.8186$$
• $$\text{normalcdf}(-10^{99},50.8,36.9,13.9) = 0.8413$$
• $$\text{invNorm}(0.80,36.9,13.9) = 48.6$$

The 80 th percentile is 48.6 years.

80% of the smartphone users in the age range 13 – 55+ are 48.6 years old or less.

Use the information in Example to answer the following questions.

## Exercise $$\PageIndex{4}$$

• Find the 30 th percentile, and interpret it in a complete sentence.
• What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old and at least 0 years old?

Let $$X =$$ a smart phone user whose age is 13 to 55+. $$X \sim N(36.9, 13.9)$$

$\text{normalcdf}(0,27,36.9,13.9) = 0.2342\nonumber$

## Example $$\PageIndex{5}$$

In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years respectively. Using this information, answer the following questions (round answers to one decimal place).

• Calculate the interquartile range ($$IQR$$).
• Forty percent of the ages that range from 13 to 55+ are at least what age?

$IQR = Q_{3} – Q_{1}\nonumber$

Calculate $$Q_{3} =$$ 75 th percentile and $$Q_{1} =$$ 25 th percentile.

\begin{align*} \text{invNorm}(0.75,36.9,13.9) &= Q_{3} = 46.2754 \\[4pt] \text{invNorm}(0.25,36.9,13.9) &= Q_{1} = 27.5246 \\[4pt] IQR &= Q_{3} - Q_{1} = 18.7508 \end{align*}

Find $$k$$ where $$P(x > k) = 0.40$$ ("At least" translates to "greater than or equal to.")

$$0.40 =$$ the area to the right.

Area to the left $$= 1 – 0.40 = 0.60$$.

The area to the left of $$k = 0.60$$.

$$\text{invNorm}(0.60,36.9,13.9) = 40.4215$$.

$$k = 40.42$$.

Forty percent of the smartphone users from 13 to 55+ are at least 40.4 years.

## Exercise $$\PageIndex{5}$$

Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean $$\mu = 81$$ points and standard deviation $$\sigma = 15$$ points.

• Calculate the first- and third-quartile scores for this exam.
• The middle 50% of the exam scores are between what two values?
• $$Q_{1} =$$ 25 th percentile $$= \text{invNorm}(0.25,81,15) = 70.9$$ $$Q_{3} =$$ 75 th percentile $$= \text{invNorm}(0.75,81,15) = 91.1$$
• The middle 50% of the scores are between 70.9 and 91.1.

## Example $$\PageIndex{6}$$

A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm.

• Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph.
• The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.
• Find the 90 th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence.

a. $$\text{normalcdf}(6,10^{99},5.85,0.24) = 0.2660$$

$$1 – 0.20 = 0.80$$

The tails of the graph of the normal distribution each have an area of 0.40.

Find $$k1$$, the 40 th percentile, and $$k2$$, the 60 th percentile ($$0.40 + 0.20 = 0.60$$).

$$k1 = \text{invNorm}(0.40,5.85,0.24) = 5.79$$ cm

$$k2 = \text{invNorm}(0.60,5.85,0.24) = 5.91$$ cm

c. 6.16: Ninety percent of the diameter of the mandarin oranges is at most 6.15 cm.

## Exercise $$\PageIndex{6}$$

Using the information from Example, answer the following:

• The middle 45% of mandarin oranges from this farm are between ______ and ______.
• Find the 16 th percentile and interpret it in a complete sentence.

The middle area $$= 0.40$$, so each tail has an area of 0.30.

$$– 0.40 = 0.60$$

The tails of the graph of the normal distribution each have an area of 0.30.

Find $$k1$$, the 30 th percentile and $$k2$$, the 70 th percentile ($$0.40 + 0.30 = 0.70$$).

$$k1 = \text{invNorm}(0.30,5.85,0.24) = 5.72$$ cm

$$k2 = \text{invNorm}(0.70,5.85,0.24) = 5.98$$ cm

$$\text{normalcdf}(5,10^{99},5.85,0.24) = 0.9998$$

• “Naegele’s rule.” Wikipedia. Available online at http://en.Wikipedia.org/wiki/Naegele's_rule (accessed May 14, 2013).
• “403: NUMMI.” Chicago Public Media & Ira Glass, 2013. Available online at www.thisamericanlife.org/radi...sode/403/nummi (accessed May 14, 2013).
• “Scratch-Off Lottery Ticket Playing Tips.” WinAtTheLottery.com, 2013. Available online at www.winatthelottery.com/publi...partment40.cfm (accessed May 14, 2013).
• “Smart Phone Users, By The Numbers.” Visual.ly, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed May 14, 2013).
• “Facebook Statistics.” Statistics Brain. Available online at http://www.statisticbrain.com/facebo...tics/(accessed May 14, 2013).

The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean $$\mu$$ and the standard deviation σ . A special normal distribution, called the standard normal distribution is the distribution of z -scores. Its mean is zero, and its standard deviation is one.

## Formula Review

• Normal Distribution: $$X \sim N(\mu, \sigma)$$ where $$\mu$$ is the mean and σ is the standard deviation.
• Standard Normal Distribution: $$Z \sim N(0, 1)$$.
• Calculator function for probability: normalcdf (lower $$x$$ value of the area, upper $$x$$ value of the area, mean, standard deviation)
• Calculator function for the $$k$$ th percentile: $$k = \text{invNorm}$$ (area to the left of $$k$$, mean, standard deviation)

## Exercise $$\PageIndex{7}$$

How would you represent the area to the left of one in a probability statement?

$$P(x < 1)$$

## Exercise $$\PageIndex{8}$$

Is $$P(x < 1)$$ equal to $$P(x \leq 1)$$? Why?

Yes, because they are the same in a continuous distribution: $$P(x = 1) = 0$$

## Exercise $$\PageIndex{9}$$

How would you represent the area to the left of three in a probability statement?

## Exercise $$\PageIndex{10}$$

What is the area to the right of three?

$$1 – P(x < 3)$$ or $$P(x > 3)$$

## Exercise $$\PageIndex{11}$$

If the area to the left of $$x$$ in a normal distribution is 0.123, what is the area to the right of $$x$$?

## Exercise $$\PageIndex{12}$$

If the area to the right of $$x$$ in a normal distribution is 0.543, what is the area to the left of $$x$$?

$$1 - 0.543 = 0.457$$

Use the following information to answer the next four exercises:

$$X \sim N(54, 8)$$

## Exercise $$\PageIndex{13}$$

Find the probability that $$x > 56$$.

## Exercise $$\PageIndex{14}$$

Find the probability that $$x < 30$$.

## Exercise $$\PageIndex{15}$$

Find the 80 th percentile.

## Exercise $$\PageIndex{16}$$

Find the 60 th percentile.

## Exercise $$\PageIndex{17}$$

$$X \sim N(6, 2)$$

Find the probability that $$x$$ is between three and nine.

## Exercise $$\PageIndex{18}$$

$$X \sim N(–3, 4)$$

Find the probability that $$x$$ is between one and four.

## Exercise $$\PageIndex{19}$$

$$X \sim N(4, 5)$$

Find the maximum of $$x$$ in the bottom quartile.

## Exercise $$\PageIndex{20}$$

Use the following information to answer the next three exercise: The life of Sunshine CD players is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. Find the probability that a CD player will break down during the guarantee period.

Figure $$\PageIndex{12}$$.

• Check student’s solution.

## Exercise $$\PageIndex{21}$$

Find the probability that a CD player will last between 2.8 and six years.

Figure $$\PageIndex{13}$$.

## Exercise $$\PageIndex{22}$$

Find the 70 th percentile of the distribution for the time a CD player lasts.

Figure $$\PageIndex{14}$$.

• 0.70, 4.78 years

## Standard Normal Distribution in Math Problems

Courtesy of C.K.Taylor (author)

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The standard normal distribution , which is more commonly known as the bell curve, shows up in a variety of places. Several different sources of data are normally distributed. As a result of this fact, our knowledge about the standard normal distribution can be used in a number of applications. But we do not need to work with a different normal distribution for every application. Instead, we work with a normal distribution with a mean of 0 and a standard deviation of 1. We will look at a few applications of this distribution that are all tied to one particular problem.

Suppose that we are told that the heights of adult males in a particular region of the world are normally distributed with a mean of 70 inches and a standard deviation of 2 inches.

• Approximately what proportion of adult males are taller than 73 inches?
• What proportion of adult males are between 72 and 73 inches?
• What height corresponds to the point where 20% of all adult males are greater than this height?
• What height corresponds to the point where 20% of all adult males are less than this height?

Before continuing on, be sure to stop and go over your work. A detailed explanation of each of these problems follows below:

• We use our z -score formula to convert 73 to a standardized score. Here we calculate (73 – 70) / 2 = 1.5. So the question becomes: what is the area under the standard normal distribution for z greater than 1.5? Consulting our table of z -scores shows us that 0.933 = 93.3% of the distribution of data is less than z = 1.5. Therefore 100% - 93.3% = 6.7% of adult males are taller than 73 inches.
• Here we convert our heights to a standardized z -score. We have seen that 73 has a z score of 1.5. The z -score of 72 is (72 – 70) / 2 = 1. Thus we are looking for the area under the normal distribution for 1< z < 1.5. A quick check of the normal distribution table shows that this proportion is 0.933 – 0.841 = 0.092 = 9.2%
• Here the question is reversed from what we have already considered. Now we look up in our table to find a z -score Z * that corresponds to an area of 0.200 above. For use in our table, we note that this is where 0.800 is below. When we look at the table, we see that z * = 0.84. We must now convert this z -score to a height. Since 0.84 = (x – 70) / 2, this means that x = 71.68 inches.
• We can use the symmetry of the normal distribution and save ourselves the trouble of looking up the value z * . Instead of z * =0.84, we have -0.84 = (x – 70)/2. Thus x = 68.32 inches.

The area of the shaded region to the left of z in the diagram above demonstrates these problems. These equations represent probabilities and have numerous applications in statistics and probability.

• Standard and Normal Excel Distribution Calculations
• Using the Standard Normal Distribution Table
• What Is Normal Distribution?
• What Is the Standard Normal Distribution?
• Examples of Z-score Calculations
• Calculating Z-Scores in Statistics
• Z-Scores Worksheet
• Formula for the Normal Distribution or Bell Curve
• Bell Curve and Normal Distribution Definition
• How to Use the NORM.INV Function in Excel
• Margin of Error Formula for Population Mean
• Examples of Confidence Intervals for Means
• Calculate Probabilities with A Standard Normal Distribution Table
• An Example of a Hypothesis Test
• The Normal Approximation to the Binomial Distribution
• How to Find the Inflection Points of a Normal Distribution
• Math Article
• Standard Normal Distribution

## The Standard Normal Distribution

The standard normal distribution is one of the forms of the normal distribution. It occurs when a normal random variable has a mean equal to zero and a standard deviation equal to one. In other words, a  normal distribution with a mean 0 and standard deviation of 1 is called the standard normal distribution. Also, the standard normal distribution is centred at zero, and the standard deviation gives the degree to which a given measurement deviates from the mean.

The random variable of a standard normal distribution is known as the standard score or a z-score . It is possible to transform every normal random variable X into a z score using the following formula:

z = (X – μ) / σ

where X is a normal random variable, μ is the mean of X, and σ is the standard deviation of X. You can also find the  normal distribution formula here. In probability theory, the normal or Gaussian distribution is a very common continuous probability distribution.

## Standard Normal Distribution Table

The standard normal distribution table gives the probability of a regularly distributed random variable Z, whose mean is equivalent to 0 and the difference equal to 1, is not exactly or equal to z. The normal distribution is a persistent probability distribution. It is also called Gaussian distribution. It is pertinent for positive estimations of z only.

A standard normal distribution table is utilized to determine the region under the bend (f(z)) to discover the probability of a specified range of distribution. The normal distribution density function f(z) is called the Bell Curve since its shape looks like a bell.

What does it mean? Is that on the off chance that you need to discover the probability of a value is not exactly or more than a fixed positive z value. You can discover it by finding it on the table. This is known as area Φ.

A standard normal distribution table presents a cumulative probability linked with a particular z-score. The rows of the table represent the whole number and tenth place of the z-score. The columns of the table represent the hundredth place. The cumulative probability (from – ∞ to the z-score) arrives in the cell of the table.

For example, a part of the standard normal table is given below. To find the cumulative probability of a z-score equal to -1.21, cross-reference the row containing -1.2 of the table with the column holding 0.01. The table explains that the probability that a standard normal random variable will be less than -1.21 is 0.1131; that is, P(Z < -1.21) = 0.1131. This table is also called a z-score table .

## Area of Standard Normal Distribution

P(Z < –a)

As specified over, the standard normal distribution table just gives the probability to values, not exactly a positive z value (i.e., z values on the right-hand side of the mean). So how would we ascertain the probability beneath a negative z value (as outlined below)?

P(Z > a)

You know Φ(a), and you realize that the total area under the standard normal curve is 1 so by numerical conclusion: P(Z > a) is 1 Φ(a).

P(Z > –a)

The probability of P(Z > –a) is P(a), which is Φ(a). To comprehend this, we have to value the symmetry of the standard normal distribution curve. We are attempting to discover the region

If this area is in the region we need.

Notice this is the same size area as the area we are searching for, just we know this area, as we can get it straight from the standard normal distribution table: it is

P(Z < a). In this way, the P(Z > –a) is P(Z < a), which is Φ(a).

Probability between z values

Let us find the probability between the values of z, i.e., a and b.

Consider the graph given below:

P(Z < b) – P(Z < a) = Φ(b) – Φ(a)

P(a < Z < b) = Φ(b) – Φ(a)

Here, the values of a and b are positive.

## Standard Normal Distribution Function

The standard normal distribution function for a random variable x is given by:

$$\begin{array}{l}Z = \frac{X-\mu_{x}}{\sigma_{x}}\end{array}$$

Probability Density Function is given by the formula,

$$\begin{array}{l}\large \varphi(x)=\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}\end{array}$$

This is a special case when μ = 0 and σ = 1. This situation of a normal distribution is also called the standard normal distribution or unit normal distribution.

Cumulative Distribution Function

The cumulative distribution function (CDF) of the standard normal distribution is generally denoted with the capital Greek letter Φ and is given by the formula:

$$\begin{array}{l}\large \Phi (x)=\frac{1}{\sqrt{2\pi}} \int_{-\infty }^{x}e^{\frac{-t^2}{2}}dt\end{array}$$

## Standard Normal Distribution Uses

• The standard normal distribution is a tool to translate a normal distribution into numbers. We may use it to get more information about the data set than was initially known.
• Standard normal distribution allows us to quickly estimate the probability of specific values befalling in our distribution or compare data sets with varying means and standard deviations.
• Also, the z-score of the standard normal distribution is interpreted as the number of standard deviations a data point falls above or below the mean.

## Characteristics of Standard Normal Distribution

A z-score of a standard normal distribution is a standard score that indicates how many standard deviations are away from the mean an individual value (x) lies:

• When z-score is positive, the x-value is greater than the mean
• When z-score is negative, the x-value is less than the mean
• When z-score is equal to 0, the x-value is equal to the mean

The empirical rule, or the 68-95-99.7 rule of standard normal distribution, tells us where most values lie in the given normal distribution. Thus, for the standard normal distribution, 68% of the observations lie within 1 standard deviation of the mean; 95% lie within two standard deviations of the mean; 99.7% lie within 3 standard deviations of the mean.

## Normal Distribution-Real World Example

Usually, happenings in the real world follow a normal distribution. This enables researchers to practice normal distribution as a model for evaluating probabilities linked with real-world scenarios. Basically, the analysis includes two steps:

• Converting raw data into the form of z-score, using the conversion equation given as z = (X – μ) / σ.
• Finding the probability. After the raw data is transformed into z-scores, then with the help of standard normal distribution tables or normal distribution calculators available online to find probabilities linked with the z-scores.

## Standard Normal Distribution Problem and Solution

Problem 1: For some computers, the time period between charges of the battery is normally distributed with a mean of 50 hours and a standard deviation of 15 hours. Rohan has one of these computers and needs to know the probability that the time period will be between 50 and 70 hours.

Solution: Let x be the random variable that represents the time period.

Given Mean, μ= 50

and standard deviation, σ = 15

To find: Pprobability that x is between 50 and 70 or P( 50< x < 70)

By using the transformation equation, we know;

For x = 50 , z = (50 – 50) / 15 = 0

For x = 70 , z = (70 – 50) / 15 = 1.33

From the table we get the value, such as;

P( 0< z < 1.33) = 0.9082 – 0.5 = 0.4082

The probability that Rohan’s computer has a time period between 50 and 70 hours is equal to 0.4082.

Problem 2: The speeds of cars are measured using a radar unit, on a motorway. The speeds are normally distributed with a mean of 90 km/hr and a standard deviation of 10 km/hr. What is the probability that a car selected at chance is moving at more than 100 km/hr?

Solution: Let the speed of cars is represented by a random variable ‘x’.

Now, given mean, μ = 90 and standard deviation, σ = 10.

To find: Probability that x is higher than 100 or P(x > 100)

For x = 100 , z = (100 – 90) / 10 = 1

P(z > 1)  = 1 – 0.8413 = 0.1587

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

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## Normal Distribution

Normal Distribution is the most common or normal form of distribution of Random Variables, hence the name “normal distribution.” It is also called Gaussian Distribution in Statistics or Probability. We use this distribution to represent a large number of random variables.

Let’s learn about Normal Distribution in detail, including its formula, characteristics, and examples.

Table of Content

## Normal Distribution Definition

Normal distribution formula, normal distribution curve, normal distribution standard deviation, normal distribution table, properties of normal distribution.

We define Normal Distribution as the probability density function of any continuous random variable for any given system. Now for defining Normal Distribution suppose we take f(x) as the probability density function for any random variable X.

Also, the function is integrated between the interval, (x, {x + dx}) then,

f(x) ≥ 0 ∀ x ϵ (−∞,+∞),

-∞ ∫ +∞ f(x) = 1

We observe that curve traced by the upper values of the Normal Distribution is in shape of a Bell, hence Normal Distribution is also called “ Bell Curve” .

## Normal Distribution Examples

We can draw Normal Distribution for various types of data that include,

• Distribution of Height of People
• Distribution of Errors in any Measurement
• Distribution of Blood Pressure of any Patient, etc.

Formula for probability density function of Normal Distribution (Gaussian Distribution) is added below,

• x is Random Variable
• σ is Standard Deviation

In any Normal Distribution, random variables are those variables that take unknown values related to the distribution and are generally bound by a range. An example of the random variable is, suppose take a distribution of the height of students in a class then the random variable can take any value in this case but is bound by a boundary of 2 ft to 6 ft, as it is generally forced physically.

Range of any normal distribution can be infinite in this case we say that normal distribution is not bothered by its range. In this case, range is extended from –∞ to + ∞.

Bell Curve still exist, in that case, all the variables in that range are called Continuous variable and their distribution is called Normal Distribution as all the values are generally closed aligned to the mean value. The graph or the curve for the same is called the Normal Distribution Curve Or Normal Distribution Graph.

We know that mean of any data spread out as a graph helps us to find the line of the symmetry of the graph whereas, Standard Deviation tells us how far the data is spread out from the mean value on either side. For smaller values of the standard deviation, the values in the graph come closer and the graph becomes narrower. While for higher values of the standard deviation the values in the graph are dispersed more and the graph becomes wider.

## Empirical Rule of Standard Deviation

Generally, the normal distribution has a positive standard deviation and the standard deviation divides the area of the normal curve into smaller parts and each part defines the percentage of data that falls into a specific region This is called the Empirical Rule of Standard Deviation in Normal Distribution.

Empirical Rule states that,

• 68% of the data approximately fall within one standard deviation of the mean, i.e. it falls between {Mean – One Standard Deviation, and Mean + One Standard Deviation}
• 95% of the data approximately fall within two standard deviations of the mean, i.e. it falls between {Mean – Two Standard Deviation, and Mean + Two Standard Deviation}
• 99.7% of the data approximately fall within a third standard deviation of the mean, i.e. it falls between {Mean – Third Standard Deviation, and Mean + Third Standard Deviation}

Studying the graph it is clear that using Empirical Rule we distribute data broadly in three parts. And thus, empirical rule is also called “68 – 95 – 99.7” rule.

Normal Distribution Table which is also called, Normal Distribution Z Table is the table of z-value for normal distribution. This Normal Distribution Z Table is given as follows:

Some important properties of normal distribution are,

• For normal distribution of data, mean, median, and mode are equal, (i.e., Mean = Median = Mode).
• Total area under the normal distribution curve is equal to 1.
• Normally distributed curve is symmetric at the center along the mean.
• In a normally distributed curve, there is exactly half value to the right of the central and exactly half value to the right side of the central value.
• Normal distribution is defined using the values of the mean and standard deviation.
• Normal distribution curve is a Unimodal Curve, i.e. a curve with only one peak.
Poisson Distribution Binomial Distribution Probability Distribution

## Normal Distribution Problems and Solutions

Let’s solve some problems on Normal Distribution

Example 1: Find the probability density function of the normal distribution of the following data. x = 2, μ = 3 and σ = 4.

Given, Variable (x) = 2 Mean = 3 Standard Deviation = 4 Using formula of probability density of normal distribution Simplifying, f(2, 3, 4) = 0.09666703

Example 2: If the value of the random variable is 4, the mean is 4 and the standard deviation is 3, then find the probability density function of the Gaussian distribution.

Given, Variable (x) = 4 Mean = 4 Standard Deviation = 3 Using formula of probability density of normal distribution Simplifying, f(4, 4, 3) = 1/(3√2π)e 0 f(4, 4, 3) = 0.13301

## FAQs on Normal Distribution

What is normal distribution.

In statistics, the normal distribution is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.

## Why is Normal Distribution Called “Normal?”

Normal Distribution also called the Gaussian Distribution is called “Normal” because it is shown that various natural processes normally follow the Gaussian distribution and hence the name “Normal Distribution”.

## What is Normal Distribution Graph?

A normal distribution graph, also known as a Gaussian distribution or bell curve, is a specific type of probability distribution. It is characterized by its symmetric, bell-shaped curve when plotted on a graph.

## What is Normal Distribution Z Table?

Z table, also known as a standard normal distribution table or a Z-score table, is a reference table used in statistics to find the probabilities associated with specific values in a standard normal distribution.

## What are characteristics of Normal Distribution?

Properties of Normal Distribution are, Normal Distribution Curve is symmetric about mean. Normal Distribution is unimodal in nature, i.e., it has single peak value. Normal Distribution Curve is always bell-shaped. Mean, Mode, and Median for Normal Distribution is always same. Normal Distribution follows Empirical Rule.

## What is Mean of Normal Distribution?

Mean (denoted as μ) represents the central or average value of data. It is also the point around which the data is symmetrically distributed.

## What is Standard Deviation of Normal Distribution?

Standard deviation (denoted as σ) measures the spread or dispersion of data points in distribution. A smaller σ indicates that data points are closely packed around mean, while a larger σ indicates more spread.

## What is Empirical Rule (68-95-99.7 Rule)?

Empirical rule for normal distribution states, Approximately 68% of data falls within one standard deviation of mean. Approximately 95% falls within two standard deviations of mean. About 99.7% falls within three standard deviations of mean.

## What are Uses of Normal Distribution?

Various uses of Normal Distribution are, For studying vrious Natural Phenomenon For studying of Financial Data. In Social Sciense for studying and predicting various parameters, etc.

## What are Limitations of Normal Distribution?

Normal Distribution is an extremely important Statical concept, but even it has some limitations such as, Various distribution of data does not follow Normal Distribution and thus it is unable to comment on these data. To much relliance of Normal Distriution or Bell curve is not a good way to prdict data as it is not 100% accurate, etc.

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## Normal distribution - practice problems

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## 5.3.3 Solved Problems

• For $0 \leq x \leq 1$, we have \begin{align}%\label{} \nonumber f_X(x)&=\int_{-\infty}^{\infty} f_{XY}(x,y)dy \\ \nonumber &=\int_{0}^{1-x}2dy\\ \nonumber &=2(1-x). \end{align} Thus, $$\nonumber f_{X}(x) = \left\{ \begin{array}{l l} 2(1-x) & \quad 0 \leq x \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right.$$ Similarly, we obtain $$\nonumber f_{Y}(y) = \left\{ \begin{array}{l l} 2(1-y) & \quad 0 \leq y \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right.$$ Thus, we have \begin{align}%\label{} \nonumber EX&=\int_{0}^{1}2x(1-x)dx\\ \nonumber &=\frac{1}{3}=EY, \end{align} \begin{align}%\label{} \nonumber EX^2&=\int_{0}^{1}2x^2(1-x)dx\\ \nonumber &=\frac{1}{6}=EY^2. \end{align} Thus, \begin{align}%\label{} \nonumber Var (X)=Var(Y)=\frac{1}{18}. \end{align} We also have \begin{align}%\label{} \nonumber EXY&=\int_{0}^{1} \int_{0}^{1-x}2xydydx\\ \nonumber &=\int_{0}^{1}x(1-x)^2dx\\ \nonumber &=\frac{1}{12}. \end{align} Now, we can find $\textrm{Cov}(X,Y)$ and $\rho(X,Y)$: \begin{align}%\label{} \nonumber \textrm{Cov}(X,Y)&=EXY-EXEY\\ \nonumber &=\frac{1}{12}-\left(\frac{1}{3}\right)^2\\ \nonumber &=-\frac{1}{36}, \end{align} \begin{align}%\label{} \nonumber \rho(X,Y)&=\frac{\textrm{Cov}(X,Y)}{\sqrt{Var(X) Var(Y)}}\\ \nonumber &=-\frac{1}{2}. \end{align}
• Note that you can look at this as a binomial experiment. In particular, we can say that $X$ and $Y$ are $Binomial(n,\frac{1}{6})$. Also, $X+Y$ is $Binomial(n,\frac{2}{6})$. Remember the variance of a $Binomial(n,p)$ random variable is $np(1-p)$. Thus, we can write \begin{align}%\label{} \nonumber n\frac{2}{6}.\frac{4}{6}&=Var(X+Y)\\ \nonumber &=Var(X)+Var(Y)+2\textrm{Cov}(X,Y)\\ \nonumber &=n\frac{1}{6}.\frac{5}{6}+n\frac{1}{6}.\frac{5}{6}+2\textrm{Cov}(X,Y). \end{align} Thus, \begin{align}%\label{} \nonumber \textrm{Cov}(X,Y)=-\frac{n}{36}. \end{align} And, \begin{align}%\label{} \nonumber \rho(X,Y)=\frac{\textrm{Cov}(X,Y)}{\sqrt{Var(X) Var(Y)}}=-\frac{1}{5}. \end{align}
• We have \begin{align}%\label{} \nonumber \textrm{Var}(U+V)&=\textrm{Var}(U)+\textrm{Var}(V)+2 \textrm{Cov}(U,V)\\ \nonumber &=1+1+2\rho_{XY}. \end{align} Since $\textrm{Var}(U+V) \geq 0$, we conclude $\rho(X,Y) \geq -1$. Also, from this we conclude that \begin{align}%\label{} \nonumber \rho(-X,Y) \geq -1. \end{align} But $\rho(-X,Y)=-\rho(X,Y)$, so we conclude $\rho(X,Y) \leq 1$.
• It is useful to find the distributions of $Z$ and $W$. To find the CDF of $Z$, we can write \begin{align}%\label{} \nonumber F_Z(z)&=P(Z \leq z) \\ \nonumber &=P(\max(X,Y) \leq z)\\ \nonumber &=P\bigg((X \leq z) \textrm{ and } (Y \leq z)\bigg)\\ \nonumber &=P(X \leq z) P(Y \leq z) &(\textrm{since }X \textrm{ and }Y \textrm{ are independent})\\ \nonumber &=F_X(z)F_Y(z). \end{align} Thus, we conclude $$\nonumber F_Z(z) = \left\{ \begin{array}{l l} 0 & \quad z 1 \end{array} \right.$$ Therefore, $$\nonumber f_Z(z) = \left\{ \begin{array}{l l} 2z & \quad 0 \leq z \leq 1 \\ 0 & \quad \text{otherwise} \end{array} \right.$$ From this we obtain $EZ=\frac{2}{3}$. Note that we can find $EW$ as follows \begin{align}%\label{} \nonumber 1=E[X+Y]&=E[Z+W] \\ \nonumber &=EZ+EW\\ \nonumber &=\frac{2}{3}+EW. \end{align} Thus, $EW=\frac{1}{3}$. Nevertheless, it is a good exercise to find the CDF and PDF of $W$, too. To find the CDF of $W$, we can write \begin{align}%\label{} \nonumber F_W(w)&=P(W \leq w) \\ \nonumber &=P(\min(X,Y) \leq w)\\ \nonumber &=1-P(\min(X,Y) > w)\\ \nonumber &=1-P\bigg((X > w) \textrm{ and } (Y > w)\bigg)\\ \nonumber &=1-P(X > w) P(Y > w) &(\textrm{since }X \textrm{ and }Y \textrm{ are independent})\\ \nonumber &=1-(1-F_X(w))(1-F_Y(w))\\ \nonumber &=F_X(w)+F_Y(w)-F_X(w)F_Y(w). \end{align} Thus, $$\nonumber F_W(w) = \left\{ \begin{array}{l l} 0 & \quad w 1 \end{array} \right.$$ Therefore, $$\nonumber f_W(w) = \left\{ \begin{array}{l l} 2-2w & \quad 0 \leq w \leq 1 \\ 0 & \text{otherwise} \end{array} \right.$$ From the above PDF we can verify that $EW=\frac{1}{3}$. Now, to find $\textrm{Cov}(Z,W)$, we can write \begin{align}%\label{} \nonumber \textrm{Cov}(Z,W)&=E[ZW]-EZEW \\ \nonumber &=E[XY]-EZEW\\ \nonumber &=E[X]E[Y]-E[Z]E[W] &(\textrm{since }X \textrm{ and }Y \textrm{ are independent})\\ \nonumber &=\frac{1}{2}.\frac{1}{2}-\frac{2}{3}.\frac{1}{3}\\ \nonumber &=\frac{1}{36}. \end{align} Note that $\textrm{Cov}(Z,W)>0$ as we expect intuitively.
• Note that since $X$ and $Y$ are jointly normal, we conclude that the random variables $X+Y$ and $X-Y$ are also jointly normal. We have \begin{align}%\label{} \nonumber \textrm{Cov}(X+Y,X-Y)&=\textrm{Cov}(X,X)-\textrm{Cov}(X,Y)+\textrm{Cov}(Y,X)-\textrm{Cov}(Y,Y)\\ \nonumber &=Var(X)-Var(Y)\\ \nonumber &=0. \end{align} Since $X+Y$ and $X-Y$ are jointly normal and uncorrelated, they are independent.
• Find $P(X+Y>0)$.
• Find the constant $a$ if we know $aX+Y$ and $X+2Y$ are independent.
• Find $P(X+Y>0|2X-Y=0)$.
• Since $X$ and $Y$ are jointly normal, the random variable $U=X+Y$ is normal. We have \begin{align}%\label{} \nonumber &EU=EX+EY=-1, \end{align} \begin{align} \nonumber Var(U)&=Var(X)+Var(Y)+2 \textrm{Cov}(X,Y) \\ \nonumber &=1+4+2 \sigma_X \sigma_Y \rho(X,Y)\\ \nonumber &=5-2 \times 1\times2\times\frac{1}{2}\\ \nonumber &=3. \end{align} Thus, $U \sim N(-1,3)$. Therefore, \begin{align}%\label{} \nonumber P(U>0)=1-\Phi\left(\frac{0-(-1)}{\sqrt{3}}\right)=1-\Phi\left(\frac{1}{\sqrt{3}}\right)=0.2819 \end{align}
• Note that $aX+Y$ and $X+2Y$ are jointly normal. Thus, for them, independence is equivalent to having $\textrm{Cov}(aX+Y,X+2Y)=0$. Also, note that $\textrm{Cov}(X,Y)=\sigma_X \sigma_Y \rho(X,Y)=-1$. We have \begin{align} \nonumber \textrm{Cov}(aX+Y,X+2Y)&=a\textrm{Cov}(X,X)+2a\textrm{Cov}(X,Y)+\textrm{Cov}(Y,X)+2\textrm{Cov}(Y,Y) \\ \nonumber &=a-(2a+1)+8\\ \nonumber &=-a+7. \end{align} Thus, $a=7$.
• If we define $U=X+Y$ and $V=2X-Y$, then note that $U$ and $V$ are jointly normal. We have \begin{align} \nonumber &EU=-1, Var(U)=3,\\ \nonumber &EV=1, Var(V)=12, \end{align} and \begin{align}%\label{} \nonumber \textrm{Cov}(U,V)&=\textrm{Cov}(X+Y,2X-Y)\\ &=2\textrm{Cov}(X,X)-\textrm{Cov}(X,Y)+2\textrm{Cov}(Y,X)-\textrm{Cov}(Y,Y)\\ \nonumber &=2Var(X)+\textrm{Cov}(X,Y)-Var(Y)\\ \nonumber &=2-1-4\\ \nonumber &=-3. \end{align} Thus, \begin{align}%\label{} \nonumber \rho(U,V)&=\frac{\textrm{Cov}(U,V)}{\sqrt{Var(U) Var(V)}}\\ \nonumber &=-\frac{1}{2}. \end{align} Using Theorem 5.4, we conclude that given $V=0$, $U$ is normally distributed with \begin{align}%\label{} \nonumber &E[U|V=0]=\mu_U+ \rho(U,V) \sigma_U \frac{0-\mu_V}{\sigma_V}=-\frac{3}{4}, \\ \nonumber &Var(U|V=0)=(1-\rho_{UV}^2)\sigma^2_U=\frac{9}{4}. \end{align} Thus \begin{align}%\label{} \nonumber P(X+Y>0|2X-Y=0)&=P(U>0|V=0)\\ \nonumber &=1-\Phi\left(\frac{0-(-\frac{3}{4})}{\frac{3}{2}}\right)\\ &= 1-\Phi\left(\frac{1}{2}\right)=0.3085. \end{align}

#### IMAGES

1. Solved problems (normal distribution)

2. Year 13

3. Solving Problems Involving Normal Distribution EXAMPLE 1 (STATISTICS

4. How to solve for the normal distribution

5. Normal Distribution Example Problems

6. 12.5 Problem Solving with Normal Distribution

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1. @btechmathshub7050 Normal Distribution

2. NORMAL DISTRIBUTION & STANDARD NORMAL DISTRIBUTION || PROPERTIES

3. Chapter 6: Normal distribution worked problem (1/2)

4. Applications of Normal Distributions

5. Normal distribution applications

6. normal distribution problems 2

1. Normal Distribution Problems with Solutions

Normal Distribution Problems with Solutions Problems and applications on normal distributions are presented. The solutions to these problems are at the bottom of the page. An online normal probability calculator and an inverse normal probability calculator may be useful to check your answers. Problems with Solutions

2. 7.E: Normal Distribution (Exercises)

Use the normal distribution to approximate the binomial distribution and find the probability of getting $$15$$ to $$18$$ heads out of $$25$$ flips. ... For this problem, use the time spent with the overweight patients. ... the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National ...

3. PDF Normal Probabilites Practice Solution

Normal Probabilities Practice Problems Solution 1. Most graduate schools of business require applicants for admission to take the Graduate Management Admission Council's GMAT examination. Scores on mean of 527 and a standard deviation of 112. What is the GMAT? Normal Distribution 500 - 527 = = - 0 .24107 112 = 527

4. Normal Distribution

Learn how to use normal distributions in statistics, with examples of properties, formulas, and applications. Find out why normal distributions matter, how to calculate the standard normal distribution, and how to use the empirical rule and the central limit theorem.

5. Normal Distributions (Bell Curve): Definition, Word Problems

A normal distribution. A normal distribution, sometimes called the bell curve (or De Moivre distribution [1]), is a distribution that occurs naturally in many situations.For example, the bell curve is seen in tests like the SAT and GRE. The bulk of students will score the average (C), while smaller numbers of students will score a B or D. An even smaller percentage of students score an F or an A.

6. 6.3: Finding Probabilities for the Normal Distribution

Definition 6.3.1 6.3. 1: z-score. z = x − μ σ (6.3.1) (6.3.1) z = x − μ σ. where μ μ = mean of the population of the x value and σ σ = standard deviation for the population of the x value. The z-score is normally distributed, with a mean of 0 and a standard deviation of 1. It is known as the standard normal curve.

7. Normal distributions review (article)

Solution: Step 1: Sketch a normal curve. Step 2: The mean of 150 cm goes in the middle.

8. The Standard Normal Distribution

x - M = 1380 − 1150 = 230. Step 2: Divide the difference by the standard deviation. SD = 150. z = 230 ÷ 150 = 1.53. The z score for a value of 1380 is 1.53. That means 1380 is 1.53 standard deviations from the mean of your distribution. Next, we can find the probability of this score using a z table.

9. Normal Distribution (solutions, examples, formulas, videos)

Math Worksheets What is the Normal Distribution? Probably the most widely known and used of all distributions is the normal distribution. It fits many human characteristics, such as height, weight, speed etc. Many living things in nature, such as trees, animals and insects have many characteristics that are normally distributed.

10. Normal Probability Practice Problems and Answers

1. Scores on a particular test are normally distributed with a standard deviation of 4 and a mean of 30. What is the probability of anyone scoring less than 40? 2. Annual salaries for a large company are approximately normally distributed with a mean of $50,000 and a of$20,000. What percentage of company workers may under $40,000? 3. 11. The Normal Distribution The normal distribution is a continuous probability distribution. This has several implications for probability. The total area under the normal curve is equal to 1. And, the probability that a normal random variable X is less than positive infinity is equal to 1; that is, P (X < ∞) = 1. The probability that a normal random variable equals ... 12. Normal distribution problem: z-scores (from ck12.org) The grade is 65. Well first, you must see how far away the grade, 65 is from the mean. So 65 will be negative because its less than the mean. 65-81 is -16. Divide that by the standard deviation, which is 6.3. So -16 divided by 6.3 is -2.54, which is the z score or "the standard deviation away from the mean. 13. 6.2: Applications of the Normal Distribution The normal distribution, which is continuous, is the most important of all the probability distributions. ... mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 10 99) by pressing 1, the EE key (a 2nd key ... the California State University Affordable Learning Solutions Program, and Merlot. We also ... 14. Normal Distribution & Probability Problems This calculus video tutorial provides a basic introduction into normal distribution and probability. It explains how to solve normal distribution problems u... 15. Normal Distribution (Definition, Formula, Table, Curve, Properties Problems and Solutions Properties Applications FAQs Normal Distribution Definition The Normal Distribution is defined by the probability density function for a continuous random variable in a system. Let us say, f (x) is the probability density function and X is the random variable. 16. PDF Solving Problems Involving Using Normal Distribution Solving Problems Involving Using Normal Distribution Problem 1: Suppose that the data concerning the first-year salaries of Baruch graduates is normally distributed with the population mean μ =$60000 and the population standard deviation σ = $15000. Find the probability of a randomly selected Baruch graduate earning less than$45000 annually.

17. Standard Normal Distribution in Math Problems

A detailed explanation of each of these problems follows below: We use our z -score formula to convert 73 to a standardized score. Here we calculate (73 - 70) / 2 = 1.5. So the question becomes: what is the area under the standard normal distribution for z greater than 1.5?

18. Standard Normal Distribution

z = (X - μ) / σ where X is a normal random variable, μ is the mean of X, and σ is the standard deviation of X. You can also find the normal distribution formula here. In probability theory, the normal or Gaussian distribution is a very common continuous probability distribution. Also, read: Probability Distribution Probability Distribution Formula

19. Normal Distribution Problems and Solutions: Step by Step

Normal Distribution Problems and Solutions: Step by Step - YouTube © 2023 Google LLC This is a great review of the Normal Distribution curve. This video assumes you know the basics. In this...

20. PDF Chapter 8 The Normal Distribution 8 THE NORMAL DISTRIBUTION

8 THE NORMAL DISTRIBUTION Objectives After studying this chapter you should appreciate the wide variety of circumstances in which the normal distribution can be used; be able to use tables of the normal distribution to solve problems; be able to use the normal distribution as an approximation to other distributions in appropriate circumstances.

21. Normal Distribution| Definition, Formula, Properties, Uses

Hermitian Matrix Normal Distribution Read Normal Distribution is the most common or normal form of distribution of Random Variables, hence the name "normal distribution." It is also called Gaussian Distribution in Statistics or Probability. We use this distribution to represent a large number of random variables.

22. Normal distribution

Lengths of 130 tubes are measured. The arithmetic mean is 17.27 cm, and the standard deviation is 1.2 cm. How many tubes do have a length: a) between 16.5 cm and 18.1 cm b) greater than 17 cm. The profit 3. An investment's profit (or loss) is normally distributed with a mean of $11,200 and a standard deviation of$8,250.

23. Solved Problems

Solution Problem I roll a fair die n n times. Let X X be the number of 1 1 's that I observe and let Y Y be the number of 2 2 's that I observe. Find Cov(X, Y) Cov ( X, Y) and ρ(X, Y) ρ ( X, Y). Hint: One way to solve this problem is to look at Var(X + Y) V a r ( X + Y) . Solution Problem