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Central Limit Theorem | Formula, Definition & Examples

Published on July 6, 2022 by Shaun Turney . Revised on June 22, 2023.

The central limit theorem states that if you take sufficiently large samples from a population, the samples’ means will be normally distributed , even if the population isn’t normally distributed.

Central Limit Theorem

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What is the central limit theorem, central limit theorem formula, sample size and the central limit theorem, conditions of the central limit theorem, importance of the central limit theorem, central limit theorem examples, practice questions, other interesting articles, frequently asked questions about the central limit theorem.

The central limit theorem relies on the concept of a sampling distribution , which is the probability distribution of a statistic for a large number of samples taken from a population.

Imagining an experiment may help you to understand sampling distributions:

  • Suppose that you draw a random sample from a population and calculate a statistic for the sample, such as the mean.
  • Now you draw another random sample of the same size, and again calculate the mean .
  • You repeat this process many times, and end up with a large number of means, one for each sample.

The distribution of the sample means is an example of a sampling distribution.

The central limit theorem says that the sampling distribution of the mean will always be normally distributed , as long as the sample size is large enough. Regardless of whether the population has a normal, Poisson, binomial, or any other distribution, the sampling distribution of the mean will be normal.

A normal distribution is a symmetrical, bell-shaped distribution, with increasingly fewer observations the further from the center of the distribution.

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Fortunately, you don’t need to actually repeatedly sample a population to know the shape of the sampling distribution. The parameters of the sampling distribution of the mean are determined by the parameters of the population:

  • The mean of the sampling distribution is the mean of the population.

\begin{equation*}\mu_{\bar{x}}=\mu\end{equation*}

  • The standard deviation of the sampling distribution is the standard deviation of the population divided by the square root of the sample size.

\begin{equation*}\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}}\end{equation*}

We can describe the sampling distribution of the mean using this notation:

\begin{equation*}\bar{X}\sim N (\mu,\dfrac{\sigma}{\sqrt{n}})\end{equation*}

  • X̄ is the sampling distribution of the sample means
  • ~ means “follows the distribution”
  • N is the normal distribution
  • µ is the mean of the population
  • σ is the standard deviation of the population
  • n is the sample size

The sample size ( n ) is the number of observations drawn from the population for each sample. The sample size is the same for all samples.

The sample size affects the sampling distribution of the mean in two ways.

1. Sample size and normality

The larger the sample size, the more closely the sampling distribution will follow a normal distribution .

When the sample size is small, the sampling distribution of the mean is sometimes non-normal. That’s because the central limit theorem only holds true when the sample size is “sufficiently large.”

By convention, we consider a sample size of 30 to be “sufficiently large.”

  • When n < 30 , the central limit theorem doesn’t apply. The sampling distribution will follow a similar distribution to the population. Therefore, the sampling distribution will only be normal if the population is normal.
  • When n ≥ 30 , the central limit theorem applies. The sampling distribution will approximately follow a normal distribution.

2. Sample size and standard deviations

The sample size affects the standard deviation of the sampling distribution. Standard deviation is a measure of the variability or spread of the distribution (i.e., how wide or narrow it is).

  • When n is low , the standard deviation is high. There’s a lot of spread in the samples’ means because they aren’t precise estimates of the population’s mean.
  • When n is high , the standard deviation is low. There’s not much spread in the samples’ means because they’re precise estimates of the population’s mean.

The central limit theorem states that the sampling distribution of the mean will always follow a normal distribution under the following conditions:

  • The sample size is sufficiently large . This condition is usually met if the sample size is n ≥ 30.
  • The samples are independent and identically distributed (i.i.d.) random variables . This condition is usually met if the sampling is random .
  • The population’s distribution has finite variance . Central limit theorem doesn’t apply to distributions with infinite variance, such as the Cauchy distribution. Most distributions have finite variance.

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homework central limit theorem

The central limit theorem is one of the most fundamental statistical theorems. In fact, the “central” in “central limit theorem” refers to the importance of the theorem.

Applying the central limit theorem to real distributions may help you to better understand how it works.

Continuous distribution

Suppose that you’re interested in the age that people retire in the United States. The population is all retired Americans, and the distribution of the population might look something like this:

Central Limit Theorem - Continuous-distribution

Age at retirement follows a left-skewed distribution. Most people retire within about five years of the mean retirement age of 65 years. However, there’s a “long tail” of people who retire much younger, such as at 50 or even 40 years old. The population has a standard deviation of 6 years.

Imagine that you take a small sample of the population. You randomly select five retirees and ask them what age they retired.

The mean of the sample is an estimate of the population mean. It might not be a very precise estimate, since the sample size is only 5.

Suppose that you repeat this procedure 10 times, taking samples of five retirees, and calculating the mean of each sample. This is a sampling distribution of the mean .

If you repeat the procedure many more times, a histogram of the sample means will look something like this:

Central Limit Theorem - Sampling-distribution

Although this sampling distribution is more normally distributed than the population, it still has a bit of a left skew .

Notice also that the spread of the sampling distribution is less than the spread of the population.

The central limit theorem says that the sampling distribution of the mean will always follow a normal distribution when the sample size is sufficiently large. This sampling distribution of the mean isn’t normally distributed because its sample size isn’t sufficiently large.

Now, imagine that you take a large sample of the population. You randomly select 50 retirees and ask them what age they retired.

The mean of the sample is an estimate of the population mean. It’s a precise estimate, because the sample size is large.

Again, you can repeat this procedure many more times, taking samples of fifty retirees, and calculating the mean of each sample:

Central Limit Theorem - Mean-of-a-large-sample

In the histogram, you can see that this sampling distribution is normally distributed, as predicted by the central limit theorem.

The standard deviation of this sampling distribution is 0.85 years, which is less than the spread of the small sample sampling distribution, and much less than the spread of the population. If you were to increase the sample size further, the spread would decrease even more.

We can use the central limit theorem formula to describe the sampling distribution:

\bar{X} \sim N (\mu,\dfrac{\sigma}{\sqrt{n}})

Discrete distribution

Approximately 10% of people are left-handed. If we assign a value of 1 to left-handedness and a value of 0 to right-handedness, the probability distribution of left-handedness for the population of all humans looks like this:

Central Limit Theorem - Theorem-discrete-distribution

The population mean is the proportion of people who are left-handed (0.1). The population standard deviation is 0.3.

Imagine that you take a random sample of five people and ask them whether they’re left-handed.

Imagine you repeat this process 10 times, randomly sampling five people and calculating the mean of the sample. This is a sampling distribution of the mean .

If you repeat this process many more times, the distribution will look something like this:

Central Limit Theorem - Theorem-discrete-distribution

The sampling distribution isn’t normally distributed because the sample size isn’t sufficiently large for the central limit theorem to apply.

As the sample size increases, the sampling distribution looks increasingly similar to a normal distribution, and the spread decreases:

Central Limit Theorem - n=10

The sampling distribution of the mean for samples with n = 30 approaches normality. When the sample size is increased further to n = 100, the sampling distribution follows a normal distribution.

We can use the central limit theorem formula to describe the sampling distribution for n = 100.

\bar{X} \sim N (0.1,\dfrac{0.3}{\sqrt{100}})

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.

  • Confidence interval
  • 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.

Normal distribution

The three types of skewness are:

  • Right skew (also called positive skew ) . A right-skewed distribution is longer on the right side of its peak than on its left.
  • Left skew (also called negative skew). A left-skewed distribution is longer on the left side of its peak than on its right.
  • Zero skew. It is symmetrical and its left and right sides are mirror images.

Skewness of a distribution

Samples are used to make inferences about populations . Samples are easier to collect data from because they are practical, cost-effective, convenient, and manageable.

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5.7: Chapter 5 Homework

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5.3 Using the Central Limit Theorem

Use the following information to answer the next eight exercises: The length of time a particular smartphone's battery lasts follows an normal distribution with a mean of ten months and a standard deviation of 10 months. A sample of 64 of these smartphones is taken.

  • What is the mean of the sampling distribution?
  • What is the standard deviation of the sampling distribution?

2 . What is the distribution for the length of time one battery lasts?

3 . What is the distribution for the mean length of time 64 batteries last?

4 . What is the distribution for the total length of time 64 batteries last?

5 . Find the probability that the sample mean is between seven and 11.

6 . Find the 80 th percentile for the total length of time 64 batteries last.

7 . Find the \(IQR\) for the mean amount of time 64 batteries last.

8 . Find the middle 80% for the total amount of time 64 batteries last.

9 . A population has a mean of 25 and a standard deviation of 2. If it is sampled repeatedly with samples of size 49, what is the mean and standard deviation of the sample means?

10 . A population has a mean of 48 and a standard deviation of 5. If it is sampled repeatedly with samples of size 36, what is the mean and standard deviation of the sample means?

11 . A population has a mean of 90 and a standard deviation of 6. If it is sampled repeatedly with samples of size 64, what is the mean and standard deviation of the sample means?

12 . A population has a mean of 120 and a standard deviation of 2.4. If it is sampled repeatedly with samples of size 40, what is the mean and standard deviation of the sample means?

13 . A population has a mean of 17 and a standard deviation of 1.2. If it is sampled repeatedly with samples of size 50, what is the mean and standard deviation of the sample means?

14 . A population has a mean of 17 and a standard deviation of 0.2. If it is sampled repeatedly with samples of size 16, what is the expected value of the mean and the standard deviation of the sample means?

15 . A population has a mean of 38 and a standard deviation of 3. If it is sampled repeatedly with samples of size 48, what is the expected value of the mean and the standard deviation of the sample means?

16 . A population has a mean of 14 and a standard deviation of 5. If it is sampled repeatedly with samples of size 60, what is the expected value of the mean and the standard deviation of the sample means?

17 . A fishing boat has 1,000 fish on board, with an average weight of 120 pounds and a standard deviation of 6.0 pounds. If sample sizes of 50 fish are checked, what is the probability the fish in a sample will have mean weight within 2.8 pounds of the true mean of the population?

18 . An experimental garden has 500 sunflower plants. The plants are being treated so they grow to unusual heights. The average height is 9.3 feet with a standard deviation of 0.5 foot. If sample sizes of 60 plants are taken, what is the probability the plants in a given sample will have an average height within 0.1 foot of the true mean of the population?

19 . A company has 800 employees. The average number of workdays between absence for illness is 123 with a standard deviation of 14 days. Samples of 50 employees are examined. What is the probability a sample has a mean of workdays with no absence for illness of at least 124 days?

20 . Cars pass an automatic speed check device that monitors 2,000 cars on a given day. This population of cars has an average speed of 67 miles per hour with a standard deviation of 2 miles per hour. If samples of 30 cars are taken, what is the probability a given sample will have an average speed within 0.50 mile per hour of the population mean?

21 . A town keeps weather records. From these records it has been determined that it rains on an average of 37% of the days each year. If 30 days are selected at random from one year, what is the probability that at least 5 and at most 11 days had rain?

22 . A maker of yardsticks has an ink problem that causes the markings to smear on 4% of the yardsticks. The daily production run is 2,000 yardsticks. What is the probability if a sample of 100 yardsticks is checked, there will be ink smeared on at most 4 yardsticks?

23 . A school has 300 students. Usually, there are an average of 21 students who are absent. If a sample of 30 students is taken on a certain day, what is the probability that at most 2 students in the sample will be absent?

24 . A college gives a placement test to 5,000 incoming students each year. On the average 1,213 place in one or more developmental courses. If a sample of 50 is taken from the 5,000, what is the probability at most 12 of those sampled will have to take at least one developmental course?

5.2 The Central Limit Theorem for Sample Means

25 . Previously, Gettysburg College statistics students estimated that the amount of change statistics students carry is normally distributed with a mean of $0.88 and a standard deviation of $0.31. Suppose that we randomly pick 25 statistics students.

  • In words, \(X\) = ____________
  • \(X \sim\) _____(_____,_____)
  • In words, \(\overline x\) = ____________
  • \(\overline x \sim\) ______ (______, ______)
  • Find the probability that an individual had between $0.80 and $1.00. Graph the situation, and shade in the area to be determined.
  • Find the probability that the average of the 25 students was between $0.80 and $1.00. Graph the situation, and shade in the area to be determined.
  • Explain why there is a difference in part e and part f.

26 . Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls.

  • If \(\overline x\) = average distance in feet for 49 fly balls, then \(\overline x \sim\) _______(_______,_______)
  • What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for \(\overline x\). Shade the region corresponding to the probability. Find the probability.
  • Find the 80th percentile of the distribution of the average of 49 fly balls.

27 . According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers.

  • In words, \(X\) = _____________
  • In words, \(\overline x\) = _____________
  • \(\overline x \sim\) _____(_____,_____)
  • Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences.
  • Would you be surprised if one taxpayer finished his or her Form 1040 in more than 12 hours? In a complete sentence, explain why.

28 . Suppose that a category of world-class runners are known to run a marathon (26 miles) in an average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races. Let \(\overline x\) the average of the 49 races.

  • Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons.
  • Find the \(80^{th}\) percentile for the average of these 49 marathons.
  • Find the median of the average running times.

29 . The length of songs in a collector’s Spotify collection is normally distributed with a mean of 2.75 minutes and a standard deviation of 0.43 minutes. Suppose we randomly pick five albums from the collection. There are a total of 43 songs on the five albums.

  • In words, \(X\) = _________
  • \(X \sim\) _____________
  • Find the first quartile for the average song length.
  • The \(IQR\) (interquartile range) for the average song length is from _______–_______.

30 . In 1940 the average size of a U.S. farm was 174 acres. Let’s say that the standard deviation was 55 acres. Suppose we randomly survey 38 farmers from 1940.

  • The \(IQR\) for \(\overline x\) is from _______ acres to _______ acres.

31 . Determine which of the following are true and which are false. Then, in complete sentences, justify your answers.

  • When the sample size is large, the mean of \(\overline x\) is approximately equal to the population mean of \(Χ\).
  • When the sample size is large, the sampling distribution of \(\overline x\) is approximately normally distributed.
  • When the sample size is large, the standard deviation of \(\overline x\) is approximately the same as the standard deviation of \(Χ\).

32 . The percent of calories from protein that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen. Let \(\overline x\) = average percent of calories from protein.

  • \(\overline x \sim\) ______(______, ______)
  • For the group of 16, find the probability that the average percent of calories from protein consumed is more than five. Graph the situation and shade in the area to be determined.
  • Find the first quartile for the average percent of calories from protein.

33 . The distribution of income in some Third World countries is considered wedge-shaped (many very poor people, very few middle income people, and even fewer wealthy people). Suppose we pick a country with a wedge-shaped distribution. Let the average salary be $2,000 per year with a standard deviation of $8,000. We randomly survey 1,000 residents of that country.

  • How is it possible for the standard deviation to be greater than the average?
  • Why is it more likely that the average of the 1,000 residents will be from $2,000 to $2,100 than from $2,100 to $2,200?

34 . Which of the following is NOT TRUE about the distribution for averages?

  • The mean, median, and mode are equal.
  • The area under the curve is one.
  • The curve never touches the x-axis.
  • The curve is skewed to the right.

35 . The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. The distribution to use for the average cost of gasoline for the 16 gas stations is:

a. \(\overline x \sim N(4.59, 0.10)\)

b. \(\overline x \sim N\left(4.59, \frac{0.10}{\sqrt{16}}\right)\)

c. \(\overline x \sim N\left(4.59, \frac{16}{0.10}\right)\)

d. \(\overline x \sim N\left(4.59, \frac{\sqrt{16}}{0.10}\right)\)

36 . A large population of 5,000 students take a practice test to prepare for a standardized test. The population mean is 140 questions correct, and the standard deviation is 80. What size samples should a researcher take to get a distribution of means of the samples with a standard deviation of 10?

37 . A large population has skewed data with a mean of 70 and a standard deviation of 6. Samples of size 100 are taken, and the distribution of the means of these samples is analyzed.

  • Will the distribution of the means be closer to a normal distribution than the distribution of the population?
  • Will the mean of the means of the samples remain close to 70?
  • Will the distribution of the means have a smaller standard deviation?
  • What is that standard deviation?

38 . A researcher is looking at data from a large population with a standard deviation that is much too large. In order to concentrate the information, the researcher decides to repeatedly sample the data and use the distribution of the means of the samples? The first effort used sample sized of 100. But the standard deviation was about double the value the researcher wanted. What is the smallest size samples the researcher can use to remedy the problem?

39 . A researcher looks at a large set of data, and concludes the population has a standard deviation of 40. Using sample sizes of 64, the researcher is able to focus the mean of the means of the sample to a narrower distribution where the standard deviation is 5. Then, the researcher realizes there was an error in the original calculations, and the initial standard deviation is really 20. Since the standard deviation of the means of the samples was obtained using the original standard deviation, this value is also impacted by the discovery of the error. What is the correct value of the standard deviation of the means of the samples?

40 . A population has a standard deviation of 50. It is sampled with samples of size 100. What is the variance of the means of the samples?

41 . A company has 1,000 employees. The average number of workdays between absence for illness is 80 with a standard deviation of 11 days. Samples of 80 employees are examined. What is the probability a sample has a mean of workdays with no absence for illness of at least 78 days and at most 84 days?

42 . Trucks pass an automatic scale that monitors 2,000 trucks. This population of trucks has an average weight of 20 tons with a standard deviation of 2 tons. If a sample of 50 trucks is taken, what is the probability the sample will have an average weight within one-half ton of the population mean?

43 . A town keeps weather records. From these records it has been determined that it rains on an average of 12% of the days each year. If 30 days are selected at random from one year, what is the probability that at most 3 days had rain?

44 . A maker of greeting cards has an ink problem that causes the ink to smear on 7% of the cards. The daily production run is 500 cards. What is the probability that if a sample of 35 cards is checked, there will be ink smeared on at most 5 cards?

45 . A school has 500 students. Usually, there are an average of 20 students who are absent. If a sample of 30 students is taken on a certain day, what is the probability that at least 2 students in the sample will be absent?

Central Limit Theorem (CLT) Calculator

  • Central Limit Theorem (CLT)

Welcome to your comprehensive guide to the Central Limit Theorem (CLT) and our interactive CLT calculator. Statistics are an integral part of our everyday lives, shaping decisions in business, politics, and social sciences alike. Whether you're a student learning the basics, a business professional analyzing data, or simply a curious mind, a good grasp of core statistical concepts like the Central Limit Theorem is invaluable.

In this guide, we aim to simplify the Central Limit Theorem for you, explain its significance, and provide a practical tool – to help visualize and explore this concept. Let's embark on this journey into the world of statistics together.

What is the Central Limit Theorem (CLT)?

The Central Limit Theorem (CLT) is one of the most powerful and essential concepts in statistics and probability theory. It holds a significant influence over many statistical techniques, including hypothesis testing, confidence intervals, and linear regression.

The CLT asserts that when an ample number of independent and identically distributed (i.i.d.) random variables are added together, their properly normalized sum tends towards a normal distribution, irrespective of the original population distribution's shape.

The CLT can be mathematically formulated as:

Let X₁, X₂, ..., Xn be a random sample of size n — that is, a sequence of independent and identically distributed random variables drawn from distributions of expected values given by µ and standard deviations given by σ.

Consider the sample average (X̄) from this distribution:

\bar{x} = \dfrac{X₁ + X₂ + ... + Xn}{n}

The Central Limit Theorem states that as n approaches infinity, the distribution of the sum (X₁ + X₂ + ... + Xn) will approximate a normal distribution (bell-shaped curve) with mean (µ) and standard deviation \sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} (σ/√n) irrespective of the shape of the original distribution.

In practical terms, if we collect samples of size n from a population, calculate each sample's mean, and create a histogram of those means, then that histogram will look like a normal distribution.

If you're a student embarking on your journey in statistics, a researcher unraveling data, or a data scientist modeling complex problems, comprehending the Central Limit Theorem and its implications is vital.

Why is the Central Limit Theorem Important?

The power of the Central Limit Theorem lies in its universality. It is the key that enables us to make inferences about the means of different populations using the normal distribution.

Here's an example:

Suppose a factory produces light bulbs, and we know that the lifespan of these light bulbs has a mean of 1000 hours, and a standard deviation of 200 hours. If we draw a sample of 100 bulbs, what's the probability that the average lifespan of these bulbs will be more than 1050 hours?

Without the CLT, this problem would be incredibly complex to solve. With the CLT, we can approximate the sample mean to a normal distribution and easily calculate the probability.

Central Limit Theorem Calculator - Bringing the Concept to Life

Our Central Limit Theorem Calculator provides an interactive way to understand and visualize the Central Limit Theorem in action. It enables you to set a population mean, population standard deviation, sample size, and the number of samples. After processing this data, the calculator generates the sample means, demonstrating the results in a graph and a table.

Making the Most of the Central Limit Theorem Calculator

Here's how you can maximize the potential of our CLT Calculator:

  • Set the population parameters: Begin by inputting the population mean and standard deviation. These values represent the overall average and variability of your data.
  • Specify your sample size: Enter the number of observations to be included in each sample. Remember, the larger the sample size, the closer the distribution of sample means will be to a normal distribution.
  • Choose the number of samples: Define the number of samples you wish to draw from the population. Each sample will be independently selected and used to compute a sample mean.
  • Analyze the results: Once you've input all the necessary parameters, our calculator will automatically make the calculations for you. The calculator will generate the distribution of sample means both as a chart and a table, also it will provide you with the standard error σx̄ (sample standard deviation) and with the sample mean X̄. Observe how this distribution becomes more normally distributed as the sample size increases. This is the power of the Central Limit Theorem in action!

Practical Applications of the Central Limit Theorem

The Central Limit Theorem is not just a theoretical concept, but it has vast practical applications:

  • Quality Control and Assurance: Industries use the Central Limit Theorem to monitor the quality of their products. By taking samples and calculating the means, they can identify any deviations from the expected quality.
  • Political Polling and Surveys: Pollsters don't survey an entire population when they're looking for statistics. They take a sample and make inferences about the population using techniques based on the Central Limit Theorem.
  • Healthcare and Pharmaceutical Research: The Central Limit Theorem plays a significant role in medical research and drug testing, helping scientists make inferences from sampled data.
  • Economics and Finance: The Central Limit Theorem is widely used for predictions in the fields of economics and finance, such as forecasting stock prices or analyzing consumer behavior.

Deepen Your Understanding with the Central Limit Theorem Calculator

The Central Limit Theorem is a cornerstone of statistics. Despite its apparent simplicity, it is a powerful tool that can be harnessed to extract meaningful insights from raw data.

Through this interactive calculator, you will not just learn the theorem but also see it at work. Seeing this principle in action will help you build a deeper, more intuitive understanding of this essential statistical theorem. So, start using the Central Limit Theorem Calculator today and take your statistical knowledge to the next level.

Remember, statistics is not just about numbers. It's about understanding the story that those numbers tell. And the Central Limit Theorem is one of the critical tools that help us decipher that story. So go ahead, start your journey towards mastering statistics with our Central Limit Theorem Calculator today.

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The Central Limit Theorem for Sample Means

Previously, De Anza statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students.

  • In words, \(Χ\) = ____________
  • \(Χ \sim\) _____(_____,_____)
  • In words, \(\overline X\) = ____________
  • \(\overline X \sim\) ______ (______, ______)
  • Find the probability that an individual had between $0.80 and $1.00. Graph the situation, and shade in the area to be determined.
  • Find the probability that the average of the 25 students was between $0.80 and $1.00. Graph the situation, and shade in the area to be determined.
  • Explain why there is a difference in part e and part f.
  • \(Χ\) = amount of change students carry
  • \(Χ \sim E(0.88, 0.88)\)
  • \(\overline X\) = average amount of change carried by a sample of 25 students.
  • \(\overline X \sim N(0.88, 0.176)\)
  • The distributions are different. Part 1 is exponential and part 2 is normal.

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls.

  • If \(\overline X\) = average distance in feet for 49 fly balls, then \(\overline X \sim\) _______(_______,_______)
  • What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for \(\overline X\). Shade the region corresponding to the probability. Find the probability.
  • Find the 80th percentile of the distribution of the average of 49 fly balls.

According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers.

  • In words, \(Χ =\) _____________
  • In words, \(\overline X\) = _____________
  • \(\overline X \sim\) _____(_____,_____)
  • Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences.
  • Would you be surprised if one taxpayer finished his or her Form 1040 in more than 12 hours? In a complete sentence, explain why.

Suppose that a category of world-class runners are known to run a marathon (26 miles) in an average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races. Let \(\overline X\) the average of the 49 races.

  • Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons.
  • Find the \(80^{th}\) percentile for the average of these 49 marathons.
  • Find the median of the average running times.

The length of songs in a collector’s iTunes album collection is uniformly distributed from two to 3.5 minutes. Suppose we randomly pick five albums from the collection. There are a total of 43 songs on the five albums.

  • In words, \(Χ\) = _________
  • \(Χ \sim\) _____________
  • Find the first quartile for the average song length.
  • The \(IQR\) (interquartile range) for the average song length is from _______–_______.

In 1940 the average size of a U.S. farm was 174 acres. Let’s say that the standard deviation was 55 acres. Suppose we randomly survey 38 farmers from 1940.

  • In words, \(Χ\) = _____________
  • The \(IQR\) for \(\overline X\) is from _______ acres to _______ acres.

Determine which of the following are true and which are false. Then, in complete sentences, justify your answers.

  • When the sample size is large, the mean of \(\overline X\) is approximately equal to the mean of \(Χ\).
  • When the sample size is large, \(\overline X\) is approximately normally distributed.
  • When the sample size is large, the standard deviation of \(\overline X\) is approximately the same as the standard deviation of \(Χ\).

The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen. Let \(\overline X\) = average percent of fat calories.

  • \(\overline X \sim\) ______(______, ______)
  • For the group of 16, find the probability that the average percent of fat calories consumed is more than five. Graph the situation and shade in the area to be determined.
  • Find the first quartile for the average percent of fat calories.

The distribution of income in some Third World countries is considered wedge shaped (many very poor people, very few middle income people, and even fewer wealthy people). Suppose we pick a country with a wedge shaped distribution. Let the average salary be $2,000 per year with a standard deviation of $8,000. We randomly survey 1,000 residents of that country.

  • How is it possible for the standard deviation to be greater than the average?
  • Why is it more likely that the average of the 1,000 residents will be from $2,000 to $2,100 than from $2,100 to $2,200?

Which of the following is NOT TRUE about the distribution for averages?

  • The mean, median, and mode are equal.
  • The area under the curve is one.
  • The curve never touches the x-axis.
  • The curve is skewed to the right.

The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. The distribution to use for the average cost of gasoline for the 16 gas stations is:

a. \(\overline X \sim N(4.59, 0.10)\)

b. \(\overline X \sim N\left(4.59, \frac{0.10}{\sqrt{16}}\right)\)

c. \(\overline X \sim N\left(4.59, \frac{16}{0.10}\right)\)

d. \(\overline X \sim N\left(4.59, \frac{\sqrt{16}}{0.10}\right)\)

Using the Central Limit Theorem

A large population of 5,000 students take a practice test to prepare for a standardized test. The population mean is 140 questions correct, and the standard deviation is 80. What size samples should a researcher take to get a distribution of means of the samples with a standard deviation of 10?

A large population has skewed data with a mean of 70 and a standard deviation of 6. Samples of size 100 are taken, and the distribution of the means of these samples is analyzed.

  • Will the distribution of the means be closer to a normal distribution than the distribution of the population?
  • Will the mean of the means of the samples remain close to 70?
  • Will the distribution of the means have a smaller standard deviation?
  • What is that standard deviation?

A researcher is looking at data from a large population with a standard deviation that is much too large. In order to concentrate the information, the researcher decides to repeatedly sample the data and use the distribution of the means of the samples? The first effort used sample sized of 100. But the standard deviation was about double the value the researcher wanted. What is the smallest size samples the researcher can use to remedy the problem?

A researcher looks at a large set of data, and concludes the population has a standard deviation of 40. Using sample sizes of 64, the researcher is able to focus the mean of the means of the sample to a narrower distribution where the standard deviation is 5. Then, the researcher realizes there was an error in the original calculations, and the initial standard deviation is really 20. Since the standard deviation of the means of the samples was obtained using the original standard deviation, this value is also impacted by the discovery of the error. What is the correct value of the standard deviation of the means of the samples?

A population has a standard deviation of 50. It is sampled with samples of size 100. What is the variance of the means of the samples?

The Central Limit Theorem for Proportions

A farmer picks pumpkins from a large field. The farmer makes samples of 260 pumpkins and inspects them. If one in fifty pumpkins are not fit to market and will be saved for seeds, what is the standard deviation of the mean of the sampling distribution of sample proportions?

A store surveys customers to see if they are satisfied with the service they received. Samples of 25 surveys are taken. One in five people are unsatisfied. What is the variance of the mean of the sampling distribution of sample proportions for the number of unsatisfied customers? What is the variance for satisfied customers?

A company gives an anonymous survey to its employees to see what percent of its employees are happy. The company is too large to check each response, so samples of 50 are taken, and the tendency is that three-fourths of the employees are happy. For the mean of the sampling distribution of sample proportions, answer the following questions, if the sample size is doubled.

  • How does this affect the mean?
  • How does this affect the standard deviation?
  • How does this affect the variance?

A pollster asks a single question with only yes and no as answer possibilities. The poll is conducted nationwide, so samples of 100 responses are taken. There are four yes answers for each no answer overall. For the mean of the sampling distribution of sample proportions, find the following for yes answers.

  • The expected value.
  • The standard deviation.
  • The variance.

The mean of the sampling distribution of sample proportions has a value of \(p\) of 0.3, and sample size of 40.

  • Is there a difference in the expected value if \(p\) and \(q\) reverse roles?
  • Is there a difference in the calculation of the standard deviation with the same reversal?

Finite Population Correction Factor

A company has 1,000 employees. The average number of workdays between absence for illness is 80 with a standard deviation of 11 days. Samples of 80 employees are examined. What is the probability a sample has a mean of workdays with no absence for illness of at least 78 days and at most 84 days?

Trucks pass an automatic scale that monitors 2,000 trucks. This population of trucks has an average weight of 20 tons with a standard deviation of 2 tons. If a sample of 50 trucks is taken, what is the probability the sample will have an average weight within one-half ton of the population mean?

A town keeps weather records. From these records it has been determined that it rains on an average of 12% of the days each year. If 30 days are selected at random from one year, what is the probability that at most 3 days had rain?

A maker of greeting cards has an ink problem that causes the ink to smear on 7% of the cards. The daily production run is 500 cards. What is the probability that if a sample of 35 cards is checked, there will be ink smeared on at most 5 cards?

A school has 500 students. Usually, there are an average of 20 students who are absent. If a sample of 30 students is taken on a certain day, what is the probability that at least 2 students in the sample will be absent?

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Mathematics > Probability

Title: on the central limit theorem for the log-partition function of 2d directed polymers.

Abstract: The log-partition function $ \log W_N(\beta)$ of the two-dimensional directed polymer in random environment is known to converge in distribution to a normal distribution when considering temperature in the subcritical regime $\beta=\beta_N=\hat{\beta}\sqrt{\pi/\log N}$, $\hat{\beta}\in (0,1)$ (Caravenna, Sun, Zygouras, Ann. Appl. Prob. (2017)). In this paper, we present an elementary proof of this result relying on a decoupling argument and the central limit theorem for sums of independent random variables. The argument is inspired by an analogy of the model to branching random walks.

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Recognition

A city near Moscow and the crater Zhukovskiy on the Moon are both named in his honor.

The State Zhukovsky Prize was established in 1920 'for the best works in mathematics'.

The Russian Air Force 's engineering academy was named for him, later reorganized into the Zhukovsky – Gagarin Air Force Academy. In May 2016, Moscow 's fourth largest airport was named in his honor.

Mosfilm produced a 1950 eponymous biopic directed by Vsevolod Pudovkin with music by Vissarion Shebalin, which earned Pudovkin and Shebalin the USSR State Prize in 1951.

The Russian Central Aero-Hydrodynamic Institute and the Ukrainian National Aerospace University – Kharkiv Aviation Institute are named after him.

The Zhukovsky House is a museum dedicated to his memory

  • Joukowsky transform
  • Kutta–Joukowski theorem
  • ↑ His surname is usually romanised as Joukovsky or Joukowsky in the literature. See for example Joukowsky transform , Kutta–Joukowski theorem and so on.
  • ↑ Blackmore, John T. (1972) (in en). Ernst Mach; His Work, Life, and Influence . University of California Press. ISBN   9780520018495 . https://books.google.co.uk/books?id=5ZY9qjB34PMC&pg=PA235&lpg=PA235&dq=ernst+mach+Karl+Lueger&source=bl&ots=z8EBBGQOyI&sig=5NpRj9nh6SnltMoaXZTNygQ-ycw&hl=en&sa=X&ved=0ahUKEwjRn7-At4_YAhWnAsAKHauWCD8Q6AEIPjAH#v=onepage&q=ernst%20mach%20Karl%20Lueger&f=false . Retrieved 16 December 2017 .  
  • ↑ Gijs A.M. van Kuik, The Lanchester-Betz-Joukowsky Limit , Wind Energ. 2007; 10:289-291
  • O'Connor, John J. ; Robertson, Edmund F. , "Nikolay Yegorovich Zhukovsky" , MacTutor History of Mathematics archive , University of St Andrews , http://www-history.mcs.st-andrews.ac.uk/Biographies/Zhukovsky.html   .

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Mathematics LibreTexts

6: The Central Limit Theorems

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In this chapter, you will study means and the central limit theorem , which is one of the most powerful and useful ideas in all of statistics. There are two alternative forms of the theorem, and both alternatives are concerned with drawing finite samples size n from a population with a known mean, \(\mu\), and a known standard deviation, \(\sigma\). The first alternative says that if we collect samples of size \(n\) with a "large enough \(n\)," calculate each sample's mean, and create a histogram of those means, then the resulting histogram will tend to have an approximate normal bell shape. The second alternative says that if we again collect samples of size \(n\) that are "large enough," calculate the sum of each sample and create a histogram, then the resulting histogram will again tend to have a normal bell-shape.

  • 6.1: Prelude to the Central Limit Theorem The central limit theorem states that, given certain conditions, the arithmetic mean of a sufficiently large number of iterates of independent random variables, each with a well-defined expected value and well-defined variance, will be approximately normally distributed.
  • 6.2E: The Central Limit Theorem for Sample Means (Exercises)
  • 6.3: The Central Limit Theorem for Sample Proportions The central limit theorem tells us that for a population with any distribution, the distribution of the sums for the sample means approaches a normal distribution as the sample size increases. In other words, if the sample size is large enough, the distribution of the sums can be approximated by a normal distribution even if the original population is not normally distributed.
  • 6.4E: Using the Central Limit Theorem (Exercises)
  • 6.5: Central Limit Theorem - Pocket Change (Worksheet) A statistics Worksheet: The student will demonstrate and compare properties of the central limit theorem.
  • 6.6: Central Limit Theorem - Cookie Recipes (Worksheet) A statistics Worksheet: The student will demonstrate and compare properties of the central limit theorem.
  • 6.E: The Central Limit Theorem (Exercises) These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.

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  • 7.3 Using the Central Limit Theorem
  • Introduction
  • 1.1 Definitions of Statistics, Probability, and Key Terms
  • 1.2 Data, Sampling, and Variation in Data and Sampling
  • 1.3 Frequency, Frequency Tables, and Levels of Measurement
  • 1.4 Experimental Design and Ethics
  • 1.5 Data Collection Experiment
  • 1.6 Sampling Experiment
  • Chapter Review
  • Bringing It Together: Homework
  • 2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs
  • 2.2 Histograms, Frequency Polygons, and Time Series Graphs
  • 2.3 Measures of the Location of the Data
  • 2.4 Box Plots
  • 2.5 Measures of the Center of the Data
  • 2.6 Skewness and the Mean, Median, and Mode
  • 2.7 Measures of the Spread of the Data
  • 2.8 Descriptive Statistics
  • Formula Review
  • 3.1 Terminology
  • 3.2 Independent and Mutually Exclusive Events
  • 3.3 Two Basic Rules of Probability
  • 3.4 Contingency Tables
  • 3.5 Tree and Venn Diagrams
  • 3.6 Probability Topics
  • Bringing It Together: Practice
  • 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable
  • 4.2 Mean or Expected Value and Standard Deviation
  • 4.3 Binomial Distribution
  • 4.4 Geometric Distribution
  • 4.5 Hypergeometric Distribution
  • 4.6 Poisson Distribution
  • 4.7 Discrete Distribution (Playing Card Experiment)
  • 4.8 Discrete Distribution (Lucky Dice Experiment)
  • 5.1 Continuous Probability Functions
  • 5.2 The Uniform Distribution
  • 5.3 The Exponential Distribution
  • 5.4 Continuous Distribution
  • 6.1 The Standard Normal Distribution
  • 6.2 Using the Normal Distribution
  • 6.3 Normal Distribution (Lap Times)
  • 6.4 Normal Distribution (Pinkie Length)
  • 7.1 The Central Limit Theorem for Sample Means (Averages)
  • 7.2 The Central Limit Theorem for Sums
  • 7.4 Central Limit Theorem (Pocket Change)
  • 7.5 Central Limit Theorem (Cookie Recipes)
  • 8.1 A Single Population Mean using the Normal Distribution
  • 8.2 A Single Population Mean using the Student t Distribution
  • 8.3 A Population Proportion
  • 8.4 Confidence Interval (Home Costs)
  • 8.5 Confidence Interval (Place of Birth)
  • 8.6 Confidence Interval (Women's Heights)
  • 9.1 Null and Alternative Hypotheses
  • 9.2 Outcomes and the Type I and Type II Errors
  • 9.3 Distribution Needed for Hypothesis Testing
  • 9.4 Rare Events, the Sample, Decision and Conclusion
  • 9.5 Additional Information and Full Hypothesis Test Examples
  • 9.6 Hypothesis Testing of a Single Mean and Single Proportion
  • 10.1 Two Population Means with Unknown Standard Deviations
  • 10.2 Two Population Means with Known Standard Deviations
  • 10.3 Comparing Two Independent Population Proportions
  • 10.4 Matched or Paired Samples
  • 10.5 Hypothesis Testing for Two Means and Two Proportions
  • 11.1 Facts About the Chi-Square Distribution
  • 11.2 Goodness-of-Fit Test
  • 11.3 Test of Independence
  • 11.4 Test for Homogeneity
  • 11.5 Comparison of the Chi-Square Tests
  • 11.6 Test of a Single Variance
  • 11.7 Lab 1: Chi-Square Goodness-of-Fit
  • 11.8 Lab 2: Chi-Square Test of Independence
  • 12.1 Linear Equations
  • 12.2 Scatter Plots
  • 12.3 The Regression Equation
  • 12.4 Testing the Significance of the Correlation Coefficient
  • 12.5 Prediction
  • 12.6 Outliers
  • 12.7 Regression (Distance from School)
  • 12.8 Regression (Textbook Cost)
  • 12.9 Regression (Fuel Efficiency)
  • 13.1 One-Way ANOVA
  • 13.2 The F Distribution and the F-Ratio
  • 13.3 Facts About the F Distribution
  • 13.4 Test of Two Variances
  • 13.5 Lab: One-Way ANOVA
  • A | Review Exercises (Ch 3-13)
  • B | Practice Tests (1-4) and Final Exams
  • C | Data Sets
  • D | Group and Partner Projects
  • E | Solution Sheets
  • F | Mathematical Phrases, Symbols, and Formulas
  • G | Notes for the TI-83, 83+, 84, 84+ Calculators

It is important for you to understand when to use the central limit theorem . If you are being asked to find the probability of the mean, use the clt for the mean. If you are being asked to find the probability of a sum or total, use the clt for sums. This also applies to percentiles for means and sums.

If you are being asked to find the probability of an individual value, do not use the clt. Use the distribution of its random variable.

Examples of the Central Limit Theorem

Law of large numbers.

The law of large numbers says that if you take samples of larger and larger size from any population, then the mean x ¯ x ¯ of the sample tends to get closer and closer to μ . From the central limit theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation gets. (Remember that the standard deviation for X ¯ X ¯ is σ n σ n .) This means that the sample mean x ¯ x ¯ must be close to the population mean μ . We can say that μ is the value that the sample means approach as n gets larger. The central limit theorem illustrates the law of large numbers.

Central Limit Theorem for the Mean and Sum Examples

Example 7.8.

A study involving stress is conducted among the students on a college campus. The stress scores follow a uniform distribution with the lowest stress score equal to one and the highest equal to five. Using a sample of 75 students, find:

  • The probability that the mean stress score for the 75 students is less than two.
  • The 90 th percentile for the mean stress score for the 75 students.
  • The probability that the total of the 75 stress scores is less than 200.
  • The 90 th percentile for the total stress score for the 75 students.

Let X = one stress score.

Problems a and b ask you to find a probability or a percentile for a mean . Problems c and d ask you to find a probability or a percentile for a total or sum . The sample size, n , is equal to 75.

Since the individual stress scores follow a uniform distribution, X ~ U (1, 5) where a = 1 and b = 5 (See Continuous Random Variables for an explanation on the uniform distribution).

μ X = a + b 2 a + b 2 = 1 + 5 2 1 + 5 2 = 3

σ X = ( b – a ) 2 12 ( b – a ) 2 12 = ( 5 – 1) 2 12 ( 5 – 1) 2 12 = 1.15

For problems a. and b., let X ¯ X ¯ = the mean stress score for the 75 students. Then,

X ¯ X ¯ ∼ N ( 3,  1 .15 75 ) ( 3,  1 .15 75 )

a. Find P ( x ¯ x ¯ < 2). Draw the graph.

a. P ( x ¯ x ¯ < 2) = 0

The probability that the mean stress score is less than two is about zero.

normalcdf ( 1,2,3, 1 .15 75 ) ( 1,2,3, 1 .15 75 ) = 0

The smallest stress score is one.

b. Find the 90 th percentile for the mean of 75 stress scores. Draw a graph.

b. Let k = the 90 th precentile.

Find k , where P ( x ¯ x ¯ < k ) = 0.90.

The 90 th percentile for the mean of 75 scores is about 3.2. This tells us that 90% of all the means of 75 stress scores are at most 3.2, and that 10% are at least 3.2.

invNorm ( 0 .90,3, 1.15 75 ) ( 0 .90,3, 1.15 75 ) = 3.2

For problems c and d, let ΣX = the sum of the 75 stress scores. Then, ΣX ~ N [(75)(3), ( 75 ) ( 75 ) (1.15)]

c. Find P ( Σx < 200). Draw the graph.

c. The mean of the sum of 75 stress scores is (75)(3) = 225

The standard deviation of the sum of 75 stress scores is ( 75 ) ( 75 ) (1.15) = 9.96

P ( Σx < 200) = 0

The probability that the total of 75 scores is less than 200 is about zero.

normalcdf (75,200,(75)(3), ( 75 ) ( 75 ) (1.15)).

The smallest total of 75 stress scores is 75, because the smallest single score is one.

d. Find the 90 th percentile for the total of 75 stress scores. Draw a graph.

d. Let k = the 90 th percentile.

Find k where P ( Σx < k ) = 0.90.

The 90 th percentile for the sum of 75 scores is about 237.8. This tells us that 90% of all the sums of 75 scores are no more than 237.8 and 10% are no less than 237.8.

invNorm (0.90,(75)(3), ( 75 ) ( 75 ) (1.15)) = 237.8

Use the information in Example 7.8 , but use a sample size of 55 to answer the following questions.

  • Find P ( x ¯ x ¯ < 7).
  • Find P ( Σx > 170).
  • Find the 80 th percentile for the mean of 55 scores.
  • Find the 85 th percentile for the sum of 55 scores.

Example 7.9

Suppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract; the analyst finds that for those people who exceed the time included in their basic contract, the excess time used follows an exponential distribution with a mean of 22 minutes.

Consider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract.

Let X = the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted time allowance.

X ∼ Exp ( 1 22 ) ( 1 22 ) . From previous chapters, we know that μ = 22 and σ = 22.

Let X ¯ X ¯ = the mean excess time used by a sample of n = 80 customers who exceed their contracted time allowance.

X ¯ X ¯  ~  N ( 22,  22 80 ) ( 22,  22 80 ) by the central limit theorem for sample means

Using the clt to find probability

  • Find the probability that the mean excess time used by the 80 customers in the sample is longer than 20 minutes. This is asking us to find P( x ¯ x ¯ > 20). Draw the graph.
  • Suppose that one customer who exceeds the time limit for his cell phone contract is randomly selected. Find the probability that this individual customer's excess time is longer than 20 minutes. This is asking us to find P ( x > 20).
  • Explain why the probabilities in parts a and b are different.

Find: P ( x ¯ x ¯ > 20)

P ( x ¯ x ¯ > 20) = 0.79199 using normalcdf ( 20,1E99,22, 22 80 ) ( 20,1E99,22, 22 80 )

The probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of 80 customers who exceed their contracted time allowance.

1E99 = 10 99 and –1E99 = –10 99 . Press the EE key for E. Or just use 10 99 instead of 1E99.

Find P (x > 20). Remember to use the exponential distribution for an individual: X ~ E x p ( 1 22 ) X ~ E x p ( 1 22 ) . P ( x > 20 )  =  e ( − ( 1 22 ) ( 20 ) ) P ( x > 20 )  =  e ( − ( 1 22 ) ( 20 ) ) or e (–0.04545(20)) = 0.4029

  • P ( x > 20) = 0.4029 but P ( x ¯ x ¯ > 20) = 0.7919
  • The probabilities are not equal because we use different distributions to calculate the probability for individuals and for means.
  • When asked to find the probability of an individual value, use the stated distribution of its random variable; do not use the clt. Use the clt with the normal distribution when you are being asked to find the probability for a mean.

Using the clt to find percentiles

Let k = the 95 th percentile. Find k where P ( x ¯ x ¯ < k ) = 0.95

k = 26.0 using invNorm ( 0 .95,22 , 22 80 ) ( 0 .95,22 , 22 80 ) = 26.0

The 95 th percentile for the sample mean excess time used is about 26.0 minutes for random samples of 80 customers who exceed their contractual allowed time.

Ninety five percent of such samples would have means under 26 minutes; only five percent of such samples would have means above 26 minutes.

Use the information in Example 7.9 , but change the sample size to 144.

  • Find P (20 < x ¯ x ¯ < 30).
  • Find P ( Σx is at least 3,000).
  • Find the 75 th percentile for the sample mean excess time of 144 customers.
  • Find the 85 th percentile for the sum of 144 excess times used by customers.

Example 7.10

In the United States, someone is sexually assaulted every two minutes, on average, according to a number of studies. Suppose the standard deviation is 0.5 minutes and the sample size is 100.

  • Find the median, the first quartile, and the third quartile for the sample mean time of sexual assaults in the United States.
  • Find the median, the first quartile, and the third quartile for the sum of sample times of sexual assaults in the United States.
  • Find the probability that a sexual assault occurs on the average between 1.75 and 1.85 minutes.
  • Find the value that is two standard deviations above the sample mean.
  • Find the IQR for the sum of the sample times.
  • 50 th percentile = μ x = μ = 2
  • 25 th percentile = invNorm (0.25,2,0.05) = 1.97
  • 75 th percentile = invNorm (0.75,2,0.05) = 2.03
  • 50 th percentile = μ Σx = n ( μ x ) = 100(2) = 200
  • 25 th percentile = invNorm(0.25,200,5) = 196.63
  • 75 th percentile = invNorm(0.75,200,5) = 203.37
  • P (1.75 < x ¯ x ¯ < 1.85) = normalcdf (1.75,1.85,2,0.05) = 0.0013
  • Using the z -score equation, z  =  x ¯ – μ x ¯ σ x ¯ z  =  x ¯ – μ x ¯ σ x ¯ , and solving for x , we have x = 2(0.05) + 2 = 2.1
  • The IQR is 75 th percentile – 25 th percentile = 203.37 – 196.63 = 6.74

Try It 7.10

Based on data from the National Health Survey, women between the ages of 18 and 24 have an average systolic blood pressures (in mm Hg) of 114.8 with a standard deviation of 13.1. Systolic blood pressure for women between the ages of 18 to 24 follow a normal distribution.

  • If one woman from this population is randomly selected, find the probability that her systolic blood pressure is greater than 120.
  • If 40 women from this population are randomly selected, find the probability that their mean systolic blood pressure is greater than 120.
  • If the sample were four women between the ages of 18 to 24 and we did not know the original distribution, could the central limit theorem be used?

Example 7.11

A study was done about violence against prostitutes and the symptoms of the posttraumatic stress that they developed. The age range of the prostitutes was 14 to 61. The mean age was 30.9 years with a standard deviation of nine years.

  • In a sample of 25 prostitutes, what is the probability that the mean age of the prostitutes is less than 35?
  • Is it likely that the mean age of the sample group could be more than 50 years? Interpret the results.
  • In a sample of 49 prostitutes, what is the probability that the sum of the ages is no less than 1,600?
  • Is it likely that the sum of the ages of the 49 prostitutes is at most 1,595? Interpret the results.
  • Find the 95 th percentile for the sample mean age of 65 prostitutes. Interpret the results.
  • Find the 90 th percentile for the sum of the ages of 65 prostitutes. Interpret the results.
  • P ( x ¯ x ¯ < 35) = normalcdf (- E 99,35,30.9,1.8) = 0.9886
  • P ( x ¯ x ¯ > 50) = normalcdf (50, E 99,30.9,1.8) ≈ 0. For this sample group, it is almost impossible for the group’s average age to be more than 50. However, it is still possible for an individual in this group to have an age greater than 50.
  • P ( Σx ≥ 1,600) = normalcdf (1600,E99,1514.10,63) = 0.0864
  • P ( Σx ≤ 1,595) = normalcdf (-E99,1595,1514.10,63) = 0.9005. This means that there is a 90% chance that the sum of the ages for the sample group n = 49 is at most 1595.
  • The 95th percentile = invNorm (0.95,30.9,1.1) = 32.7. This indicates that 95% of the prostitutes in the sample of 65 are younger than 32.7 years, on average.
  • The 90th percentile = invNorm (0.90,2008.5,72.56) = 2101.5. This indicates that 90% of the prostitutes in the sample of 65 have a sum of ages less than 2,101.5 years.

Try It 7.11

According to Boeing data, the 757 airliner carries 200 passengers and has doors with a height of 72 inches. Assume for a certain population of men we have a mean height of 69.0 inches and a standard deviation of 2.8 inches.

  • What doorway height would allow 95% of men to enter the aircraft without bending?
  • Assume that half of the 200 passengers are men. What mean doorway height satisfies the condition that there is a 0.95 probability that this height is greater than the mean height of 100 men?
  • For engineers designing the 757, which result is more relevant: the height from part a or part b? Why?

HISTORICAL NOTE

: Normal Approximation to the Binomial

Historically, being able to compute binomial probabilities was one of the most important applications of the central limit theorem. Binomial probabilities with a small value for n (say, 20) were displayed in a table in a book. To calculate the probabilities with large values of n , you had to use the binomial formula, which could be very complicated. Using the normal approximation to the binomial distribution simplified the process. To compute the normal approximation to the binomial distribution, take a simple random sample from a population. You must meet the conditions for a binomial distribution :

  • there are a certain number n of independent trials
  • the outcomes of any trial are success or failure
  • each trial has the same probability of a success p

Recall that if X is the binomial random variable, then X ~ B ( n, p ). The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five ( np > 5 and nq > 5; the approximation is better if they are both greater than or equal to 10). Then the binomial can be approximated by the normal distribution with mean μ = np and standard deviation σ = n p q n p q . Remember that q = 1 – p . In order to get the best approximation, add 0.5 to x or subtract 0.5 from x (use x + 0.5 or x – 0.5). The number 0.5 is called the continuity correction factor and is used in the following example.

Example 7.12

Suppose in a local Kindergarten through 12 th grade (K - 12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed.

  • Find the probability that at least 150 favor a charter school.
  • Find the probability that at most 160 favor a charter school.
  • Find the probability that more than 155 favor a charter school.
  • Find the probability that fewer than 147 favor a charter school.
  • Find the probability that exactly 175 favor a charter school.

Let X = the number that favor a charter school for grades K trough 5. X ~ B ( n, p ) where n = 300 and p = 0.53. Since np > 5 and nq > 5, use the normal approximation to the binomial. The formulas for the mean and standard deviation are μ = np and σ = n p q n p q . The mean is 159 and the standard deviation is 8.6447. The random variable for the normal distribution is Y . Y ~ N (159, 8.6447). See The Normal Distribution for help with calculator instructions.

For part a, you include 150 so P ( X ≥ 150) has normal approximation P ( Y ≥ 149.5) = 0.8641.

normalcdf (149.5,10^99,159,8.6447) = 0.8641.

For part b, you include 160 so P ( X ≤ 160) has normal appraximation P ( Y ≤ 160.5) = 0.5689.

normalcdf (0,160.5,159,8.6447) = 0.5689

For part c, you exclude 155 so P ( X > 155) has normal approximation P ( y > 155.5) = 0.6572.

normalcdf (155.5,10^99,159,8.6447) = 0.6572.

For part d, you exclude 147 so P ( X < 147) has normal approximation P ( Y < 146.5) = 0.0741.

normalcdf (0,146.5,159,8.6447) = 0.0741

For part e, P ( X = 175) has normal approximation P (174.5 < Y < 175.5) = 0.0083.

normalcdf (174.5,175.5,159,8.6447) = 0.0083

Because of calculators and computer software that let you calculate binomial probabilities for large values of n easily, it is not necessary to use the the normal approximation to the binomial distribution, provided that you have access to these technology tools. Most school labs have Microsoft Excel, an example of computer software that calculates binomial probabilities. Many students have access to the TI-83 or 84 series calculators, and they easily calculate probabilities for the binomial distribution. If you type in "binomial probability distribution calculation" in an Internet browser, you can find at least one online calculator for the binomial.

For Example 7.12 , the probabilities are calculated using the following binomial distribution: ( n = 300 and p = 0.53). Compare the binomial and normal distribution answers. See Discrete Random Variables for help with calculator instructions for the binomial.

P ( X ≥ 150) : 1 - binomialcdf (300,0.53,149) = 0.8641

P ( X ≤ 160) : binomialcdf (300,0.53,160) = 0.5684

P ( X > 155) : 1 - binomialcdf (300,0.53,155) = 0.6576

P ( X < 147) : binomialcdf (300,0.53,146) = 0.0742

P ( X = 175) :(You use the binomial pdf.) binomialpdf (300,0.53,175) = 0.0083

Try It 7.12

In a city, 46 percent of the population favor the incumbent, Dawn Morgan, for mayor. A simple random sample of 500 is taken. Using the continuity correction factor, find the probability that at least 250 favor Dawn Morgan for mayor.

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  • Book title: Introductory Statistics
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  • Book URL: https://openstax.org/books/introductory-statistics/pages/1-introduction
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