6.3: The Central Limit Theorem

Last updated
Save as PDF

Page ID: 58280

Toros Berberyan, Tracy Nguyen, and Alfie Swan
Citrus College

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\dsum}{\displaystyle\sum\limits} $

$ \newcommand{\dint}{\displaystyle\int\limits} $

$ \newcommand{\dlim}{\displaystyle\lim\limits} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

Learning Objectives

Understand the Central Limit Theorem, stating that the distribution of sample means approaches a normal distribution as sample size increases, regardless of the. population’s shape
Use the Central Limit Theorem to compute probabilities about sample means using z-scores.
Apply this theorem in inferential statistics, including confidence intervals and hypothesis testing.

The central limit theorem (CLT) states that, regardless of the original population distribution, the sampling distribution of the sample mean will approach a normal distribution as the sample size increases, typically when the sample size is 30 or more. This theorem is fundamental in statistics because it allows for inferential analysis, meaning we can make predictions about a population using sample data. It is widely used in hypothesis testing, confidence intervals, and quality control, helping researchers and analysts conclude large populations from smaller, manageable samples.

Definition $\PageIndex{1}$:Statistical Inference

Statistical Inference: to make accurate decisions about parameters from statistics.

When it says “accurate decision,” you want to be able to measure how accurate. You measure how accurate using probability. In both binomial and normal distributions, you needed to know that the random variable followed either distribution. You need to know how the statistic is distributed, and then you can find probabilities. In other words, you need to know the shape of the sample mean or whatever statistic you want to decide about.

How is the statistic distributed? This is answered with a sampling distribution.

Definition $\PageIndex{2}$: Sampling Distribution

Sampling Distribution: how a sample statistic is distributed when repeated trials of size n are taken.

This depends on how the original distribution is distributed. However, using the Central Limit Theorem, we do not need to know the shape of the population distribution. The sampling distribution will be normal as long as n is at least 30.

Definition: Central Limit Theorem

The central limit theorem (CLT) states that, for a sufficiently large sample size, the distribution of the sample mean will be approximately normal, regardless of the shape of the original population distribution. This means that as the sample size increases, the sampling distribution of the mean becomes more normally distributed, making it useful for statistical inference, hypothesis testing, and confidence interval estimation.

The mean of the distribution will $\mu$
The standard deviation is called the standard error and is $ \dfrac{ \sigma }{\sqrt{n}} $.

What this says is that no matter what x looks like, $\overline{x}$ would look normal if n is large enough. Now, what size of n is large enough? That depends on how x is distributed in the first place. If the original random variable is normally distributed, then n just needs to be 2 or more data points. If the original random variable is somewhat mound-shaped and symmetrical, then n needs to be greater than or equal to 30. Sometimes the sample size can be smaller, but this is a good rule of thumb. The sample size may have to be much larger if the original random variable is skewed one way or another.

Now that you know what the sample mean will look like in a normal distribution, you can find the probability related to the sample mean. Remember that the mean of the sample mean is just the mean of the original data ($\mu_{\overline{x}}=\mu$ ), but the standard deviation of the sample mean, $\sigma_{\overline{x}}$, also known as the standard error of the mean, is actually $\sigma_{\overline{x}}=\dfrac{\sigma}{\sqrt{n}}$. Make sure you use this in all calculations. If you are using the z-score, the formula when working with $\overline{x}$ is $z=\dfrac{\overline{x}-\mu_{\overline{x}}}{\sigma_{\overline{x}}}=\dfrac{\overline{x}-\mu}{\sigma / \sqrt{n}}$. If you are using the TI-83/84 calculator, then the input would be normalcdf(lower limit, upper limit, $\mu$, $\sigma / \sqrt{n}$ ). If you are using R, then the input would be pnorm( $\overline{x}, \mu, \sigma / \operatorname{sqrt}(n) $) to find the area to the left of $\overline{x}$. Remember to subtract pnorm( $\overline{x}, \mu, \sigma / \operatorname{sqrt}(n))$ ) from 1 if you want the area to the right of $\overline{x}$.

Example $\PageIndex{1}$ general normal distribution

The length of a human pregnancy is normally distributed with a mean of 272 days with a standard deviation of 9 days (Bhat & Kushtagi, 2006).

Click on blue links to access the normal distribution tables for negative z-values and positive z-values. After the value is found, press "back" to return to the page.

State the random variable.
Find the probability that the mean of 30 pregnancies lasts more than 273.46 days.
Find the probability that the mean of 30 pregnancies lasts less than 267.99 days.
Find the probability that the mean of 30 pregnancies lasts between 267.99 and 273.46 days.

Solution

a. $\overline{x}$ = mean length of 30 human pregnancies

b. First, translate the statement into a mathematical statement.

$P({\overline{x}}>273.46)$

Now, draw a picture. Remember that the center of this normal curve is 272.

Description of graph see below. — Figure $\PageIndex{1}$: Normal distribution where $\mu$ of 272 is located in the middle where ${\overline{x}}$ = 273.46 is a fence on the left of the center. The shaded area is on the right.

To find the probability on the TI-83/84, looking at the picture, you realize the lower limit is 273.46. The upper limit is infinity. The calculator doesn’t have infinity on it, so you need to put in a really big number. Some people like to put in 1000, but if you are working with numbers that are bigger than 1000, then you would have to remember to change the upper limit. The safest number to use is $1 \times 10^{99}$, which you put in the calculator as 1E99 (where E is the EE button on the calculator). The $\mu$ (population mean) will be 272. The population standard deviation of our sample means ( $\sigma_{\overline{x}}$ ) will be $\dfrac{\sigma}{\sqrt{n}}$.

The work to find $\sigma_{\overline{x}}$ is provided below.

$\sigma_{\overline{x}}=\dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\overline{x}}=\dfrac{9}{\sqrt{30}}$

$\sigma_{\overline{x}}=1.64$

As a result, to find the probability, we will use the normalcdf() command in our TI 83/84 calculator where the lower limit will be 273.46, the upper limit will be 1 E 99, $\mu$ (population mean) will be 272, and population standard deviation of our sample means ( $\sigma_{\overline{x}}$ ) will be 1.64.

The command looks like:

$\text{normalcdf}(273.46,1 E 99,272,1.64)$

TI-84+ calculator output for normal function is 0.1867. — Figure $\PageIndex{2}$: TI-83/84 Output for Area to the Right of 273.46.

Thus, $P({\overline{x}}>273.46) \approx 0.1867$

Thus, 18.67% of the mean pregnancies of 30 women last more than 273.46 days. This is not unusual since the probability is greater than 5%.

To find the probability using a z-table, we need to first compute the z-score and then use the table to find the probability. Based on the problem, we know that $\mu$ (population mean) is 272 days and $\sigma$ (population standard deviation) is 9 days. We know that we are finding the probability of the mean of 30 pregnancies lasting more than 273.46 days, so $\overline{x}$ will be 273.46 days.

$z=\dfrac{\overline{x}-\mu_{\overline{x}}}{\sigma_{\overline{x}}}=\dfrac{(\overline{x}-\mu)}{\sigma / \sqrt{n}}$

$z=\dfrac{(273.46-272)}{9/ \sqrt{30}} $

$z=\dfrac{1.46}{1.64}$

$z=0.89$ (rounded to two decimal places)

As a result, $P({\overline{x}}>273.46) = P(z > 0.89)$. Since z is more than 0.89, our shaded area will be on the right of z = 0.89.

We can use the z-table to find our probability to the left. If we convert the normal distribution to the standard normal distribution, we get the following:

Normal distribution where z = 0.89 and the the shaded area is on the right. — Figure $\PageIndex{3}$: Normal Distribution, Z = 0.89, Shaded on the Right

From the z-table, it only gives us the probability to the left of the z-score. Since we know the total area of the normal distribution is 1. To get the area to the right, we need to take 1 minus the area to the left. The work is provided below.

P(z > 0.89) = 1 - P(z < 0.89)

To find the area or probability to the left of z = 0.89, we use the standard normal distribution table provided below. We go down 0.8 and move over to the right 0.09 to get the area of 0.8133. This area (probability) represents the area to the left of z=0.89.

Table $\PageIndex{5}$: Normal Distribution Table For Positive Z-values

Based on the z-table above, the Left Area will be 0.8133.

Right Area = 1 - Left Area

Right Area = 1 - 0.8133 = 0.1867

P(z>0.89)= 0.1867 = 18.67%

Using the table, $P({\overline{x}}>273.46)$ = 18.67%. This means that the probability that the mean of 30 pregnancies will last more than 273.46 days is 18.67%.

The answer using the z-table is slightly off due to rounding our z-value to two decimal places. The calculator uses more decimal places for z, hence it has a more accurate answer.

c. First, translate the statement into a mathematical statement.

$P({\overline{x}}<267.99)$

Now, draw a picture. Remember that the center of this normal curve is 272.

Normal distribution to the left of 267.99. — Figure $\PageIndex{4}$: Normal Distribution with Area to the Left of 267.99

To find the probability on the TI-83/84, looking at the picture, though it is hard to see in this case, the lower limit is negative infinity. Again, the calculator doesn’t have this on it, put in a really small number, such as $-1 \times 10^{99}=-1 E 99$ on the calculator.

TI-84+ output for normal cumulative funtion with area 0.0072. — Figure $\PageIndex{5}$: TI-83/84 Output for the Area to the Left of 267.99

$P({\overline{x}}<267.99)=\text {normalcdf }(-1 E 99,267.99,272,1.64)=0.0072$

Thus, the probability that the mean of 30 pregnancies lasts less than 267.99 days is 0.72%. This is unusual since the probability is less than 5%.

To find the probability using a z-table, we need to first compute the z-score and then use the table to find the probability. Based on the problem, we know that $\mu$ (population mean) is 272 days and $\sigma$ (population standard deviation) is 9 days. We know that we are finding the probability of the mean of 30 pregnancies that last less than 267.99 days, so ${\overline{x}}$ will be 267.99 days.

$z=\dfrac{\overline{x}-\mu_{\overline{x}}}{\sigma_{\overline{x}}}=\dfrac{(\overline{x}-\mu)}{\sigma / \sqrt{n}}$

$z=\dfrac{(267.99-272)}{9/ \sqrt{30}} $

$z=\dfrac{-4.01}{1.64}$

$z=-2.45$ (rounded to two decimal places)

As a result, $P({\overline{x}}<267.99)$ = P(z < -2.45). Since z is less than -2.45, our shaded area will be on the left of z = -2.45.

We can use the z-table to find our probability to the left. If we convert the normal distribution to the standard normal distribution, we get the following:

Figure $\PageIndex{6}$: Normal Distribution, Z = -2.45, Shaded on the Left

We use the standard normal distribution table provided below to find the area or probability to the left of z = -2.45. We go down -2.4 and move over to the right 0.05 to get the area of 0.0071. This area (probability) represents the area to the left of z=-2.45.

Table $\PageIndex{6}$: Normal Distribution Table For Negative Z-values

Based on the z-table above, the Left Area will be 0.0071, which is the probability that z is less than -2.45. If we express the answer as a percentage, we get the following answer:

P( z < -2.45) = 0.71%

Using the table, $P({\overline{x}}<267.99)$ = 0.71%. This means the probability that the mean of 30 pregnancies will last less than 267.99 days is 0.71%.

The answer using the z-table is slightly off due to rounding our z-value to two decimal places. The calculator uses more decimal places for z, hence it has a more accurate answer.

d. First translate the statement into a mathematical statement.

$P(267.99<{\overline{x}}<273.46)$

Now, draw a picture. Remember that the center of this normal curve is 272.

Normal distribution with area between 267.99 and 273.46. — Figure $\PageIndex{7}$: Normal Distribution with an Area Between 267.99 and 273.46

In this case, the lower limit is 267.99 and the upper limit is 273.46.

Using the calculator

TI-84+ Output of normal cumulative function with area of 0.8061. — Figure $\PageIndex{8}$: TI-83/84 Output for Area Between 267.99 and 273.46

$P(267.99<{\overline{x}}<273.46)=\text {normalcdf }(267.99,273.46,272,1.64)=0.8061$

Thus, the probability that the mean of 30 pregnancies will last between 267.99 and 273.46 days is 80.61%.

To find the probability using a z-table, we need to first compute the z-scores and then use the table to find the probability. Based on the problem, we know that $\mu$ (population mean) is 272 days and $\sigma$ (population standard deviation) is 9 days. We know that we are finding the probability of the mean of 30 pregnancies lasting between 267.99 and 273.46 days, so ${\overline{x}}$ will be 267.99 days and 273.46 days. For each ${\overline{x}}$, we need to find its corresponding z-value.

$z=\dfrac{\overline{x}-\mu_{\overline{x}}}{\sigma_{\overline{x}}}=\dfrac{(\overline{x}-\mu)}{\sigma / \sqrt{n}}$

$z=\dfrac{(267.99-272)}{9/ \sqrt{30}} $

$z=\dfrac{-4.01}{1.64}$

$z=-2.45$ (rounded to two decimal places)

$z=\dfrac{\overline{x}-\mu_{\overline{x}}}{\sigma_{\overline{x}}}=\dfrac{(\overline{x}-\mu)}{\sigma / \sqrt{n}}$

$z=\dfrac{(273.46-272)}{9/ \sqrt{30}} $

$z=\dfrac{1.46}{1.64}$

$z=0.89$ (rounded to two decimal places)

As a result, $P(267.99<{\overline{x}}<273.46)$ = P(-2.45 < z < 0.89).

Figure $\PageIndex{9}$: Normal Distribution Where Shaded Area is Between z = -2.45 and z = 0.89

We can find our probability to the left of each z-score by using the z-table. If we subtract the higher probability value from the lower probability value, we will find the area (probability) between z = -2.45 and z = 0.89.

$P\left(-2.45<z<0.89\right)=P\left(z<0.89\right)-P\left(z<-2.45\right)$

To find $P\left(z<0.89\right)$ and $P\left(z<-2.45\right)$ we use the z-table.

z = 0.89

$P\left(z<0.89\right)=0.8133$

Table $\PageIndex{5}$: Normal Distribution Table For Positive Z-values

z = -2.45

To find the area or probability to the left of z = -2.45, we use the standard normal distribution table provided below. We go down -2.4 and move over to the right by 0.05 to get an area of 0.0071. This area (probability) represents the area to the left of z=-2.45.

$P\left(z<-2.45\right)=0.0071$

Table $\PageIndex{6}$: Normal Distribution Table For Negative Z-values

Now that we found $P\left(z<0.89\right)=0.8133$ and $P\left(z<-2.45\right)=0.0071$, we can use the following formula:

$P(267.99<{\overline{x}}<273.46)=P\left(-2.45<z<0.89\right)=P\left(z<0.89\right)-P\left(z<-2.45\right)$

$P(267.99<{\overline{x}}<273.46)=P\left(-2.45<z<0.89\right)=0.8133-0.0071$

$P(267.99<{\overline{x}}<273.46)=P\left(-2.45<z<0.89\right)=0.8062$

Thus, the probability that the mean of 30 pregnancies will last between 267.99 and 273.46 days will be 80.62%.

Authors

"6.3: The Central Limit Theorem" by Toros Berberyan, Tracy Nguyen, and Alfie Swan is licensed under CC BY-SA 4.0

Attributions

"6.5: Sampling Distribution and the Central Limit Theorem" by Kathryn Kozak is licensed under CC BY-SA 4.0

Exercises

The mean starting salary for nurses is $67,694 nationally ("Staff nurse -," 2013). The standard deviation is approximately $10,333. The starting salary is approximately normally distributed. A sample of 42 starting salaries for nurses is taken. Find the probability that the sample mean is less than $65,000. Use the standard normal distribution table, round the z-value to two decimal places, and round the final answer to four decimal places.

Scan the QR code or click on it to open the MyOpenMath version of the above question with step-by-step guidance.
MyOpenMath is a free online learning platform designed to support math instruction through automated homework, quizzes, and assessments. You must register for MyOpenMath and sign in to view the question.

The mean starting salary for nurses is $67,694 nationally ("Staff nurse -," 2013). The standard deviation is approximately $10,333. The starting salary is approximately normally distributed. A sample of 42 starting salaries for nurses is taken. Find the probability that the sample mean is more than $70,000. Use the standard normal distribution table, round the z-value to two decimal places, and round the final answer to four decimal places.

The mean starting salary for nurses is $67,694 nationally ("Staff nurse -," 2013). The standard deviation is approximately $10,333. The starting salary is approximately normally distributed. A sample of 42 starting salaries for nurses is taken. Find the probability that the sample mean is between $66,000 and $71,000.Use the standard normal distribution table, round the z-value to two decimal places, and round the final answer to four decimal places.

The size of fish is significant to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm (Ovegard, Berndt & Lunneryd, 2012). The length of fish is normally distributed. A sample of 35 fish is taken. Find the probability that the sample mean length of the Atlantic cod is less than 52 cm. Use the standard normal distribution table, round the z-value to two decimal places, and round the final answer to four decimal places.

The size of fish is significant to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm (Ovegard, Berndt & Lunneryd, 2012). The length of fish is normally distributed. A sample of 35 fish is taken. Find the probability that the sample mean length of the Atlantic cod is more than 50.5 cm. Use the standard normal distribution table, round the z-value to two decimal places, and round the final answer to four decimal places.

The size of fish is significant to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm (Ovegard, Berndt & Lunneryd, 2012). The length of fish is normally distributed. A sample of 35 fish is taken. Find the probability that the sample mean length of the Atlantic cod is between 48.5 cm. and 50.8 cm. Use the standard normal distribution table, round the z-value to two decimal places, and round the final answer to four decimal places.

At a large community college, the time it takes for a student to transfer to a four-year university is known to be approximately normal, with a mean of 3.5 years and a standard deviation of 1.2 years. Suppose a random sample of 36 students is selected. What is the probability that the average time it takes for these students to transfer is between 2.5 and 3.3 years? Round the final answers to four decimal places.

At a large community college, the time it takes for a student to transfer to a four-year university is known to be approximately normal, with a mean of 3.5 years and a standard deviation of 1.2 years. Suppose a random sample of 36 students is selected. What is the probability that the average time it takes for these students to transfer is less than 3 years? Round the final answers to four decimal places.

At a large community college, the time it takes for a student to transfer to a four-year university is known to be approximately normal, with a mean of 3.5 years and a standard deviation of 1.2 years. Suppose a random sample of 36 students is selected. What is the probability that the average time it takes for these students to transfer is more than 3.8 years? Round the final answers to four decimal places.

Answers

If you are an instructor and want the solutions to all the exercise questions for each section, please email Toros Berberyan.

Search

Text Color

Text Size

Margin Size

Font Type

Solution

Right Area = 1 - Left Area

z = 0.89

z = -2.45

Learning Objectives

Definition \(\PageIndex{1}\):Statistical Inference

Definition \(\PageIndex{2}\): Sampling Distribution

Definition: Central Limit Theorem

Example \(\PageIndex{1}\) general normal distribution

Solution

Right Area = 1 - Left Area

z = 0.89

z = -2.45

Authors

Attributions

Exercises