Chapter 5.1, Problem 20E

(a)

Section 1:

To determine

To find: Themean for the average number of friends per user for the provided sample.

Answer to Problem 20E

Solution: The mean is 190friends.

Explanation of Solution

The sample mean of average number of friends per user is calculated by the formula as:

$\begin{array}{c}{\mu }_{\overline{x}}=\frac{{\displaystyle \sum _{i=1}^{n=70}{\overline{x}}_i}}{n}\\ =\frac{{\overline{x}}_1+{\overline{x}}_2+\mathrm{..}+{\overline{x}}_{69}+{\overline{x}}_{70}}{70}\\ =\frac{190+190+\mathrm{...}+190}{70}\\ =190\end{array}$

It is provided that average number of friends on Facebook for eachuser is 190and the sample of70adult users is taken. Hence, the mean of the average number of friends per adult user is 190friends.

Section 2:

To determine

To find: The sample standard deviationfor the mean number of friends per adult user.

Answer to Problem 20E

Solution: The sample standard deviation is $34.423$.

Explanation of Solution

The standard deviation of sample means is calculated by the formula as:

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{288}{\sqrt{70}}\\ =34.423\end{array}$

Hence, the sample standard deviation of mean of numbers friends per adult user is $34.423$.

(b)

To determine

To find: The probability that the average number of friends for 70 Facebook users is greater than 250.

Answer to Problem 20E

Solution: The probability is $0.0409$.

Explanation of Solution

The probability that the average number of friends for 70Facebook users is greater than 250will be given by:

$P\left(\overline{x}>250\right)$

The probability that $\overline{x}$ is greater than 250is the area to the right of the normal curve. To find the area, first convert the value of sample mean $\overline{x}>250$ to a z-score and then obtain the area to the right of z-score from the standard normal table provided in the book.

The z-score for the value 250is calculated as:

$\begin{array}{c}P\left(\overline{x}>250\right)=P\left(\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}>\frac{250-190}{288/\sqrt{70}}\right)\\ =P\left(z>1.74\right)\\ =1-P\left(z\le 1.74\right)\\ =1-0.9591\\ =0.0409\end{array}$

The area under the curveto the right of $P\left(z\le 1.74\right)$ by using the standard normal table is obtained as $0.0409$.

Therefore, the required probability that the sample of 70 Facebook users has minimum of 250 friends is $0.0409$.

(c)

Section 1:

To determine

To find: The mean of total number of friends in the provided sample.

Answer to Problem 20E

Solution: The mean is27040.

Explanation of Solution

The sample mean of mean follow the approximate normal distribution with mean as 190and sample standard deviation as $34.423$ and the sample mean of total number of friends is calculated as:

$\overline{x}\approx N\left(190,{34.423}^2\right)$

And the total number of friends follows normal distribution with mean and variance as::

$\begin{array}{l}{\displaystyle \sum _{i=1}^{70}{x}_i}\approx N\left(70\times 190,70\times {34.423}^2\right)\\ {\displaystyle \sum _{i=1}^{70}{x}_i}\approx N\left(13300,82946.01\right)\end{array}$

Hence, the mean of total number of friends in the sample is 13300.

Section 2:

To determine

To find: The standard deviation of total number of friends in the provided sample.

Answer to Problem 20E

Solution: The standard deviation is 288.

Explanation of Solution

The sample standard deviation of mean follow the approximate normal distribution with mean as 190and sample standard deviation as $34.423$ and the sample standard deviation of total number of friends is calculated as:

$\overline{x}\approx N\left(190,{34.423}^2\right)$

And the total number of friends follows normal distribution with mean and variance as:

$\begin{array}{l}{\displaystyle \sum _{i=1}^{70}{x}_i}\approx N\left(70\times 190,70\times {34.423}^2\right)\\ {\displaystyle \sum _{i=1}^{70}{x}_i}\approx N\left(13300,82946.01\right)\end{array}$

Hence, the standard deviation of total number of friends is calculated as:

$\sqrt{82946.01}=288$

(d)

To determine

To find: The probability that the total number of friends for 70Facebook users is greater than 17500.

Answer to Problem 20E

Solution: The probability is $1$.

Explanation of Solution

The required probability that the mean of total number of friends of 70users is greater than 17500; that is the required probability will be given by:

$\begin{array}{l}{\displaystyle \sum _{i=1}^{70}{x}_i}\approx N\left(70\times 190,70\times {34.423}^2\right)\\ y={\displaystyle \sum _{i=1}^{70}{x}_i}\approx N\left(13300,82946.01\right)\end{array}$

The probability that $y$ is greater than 17500is the area to the right of the normal curve. To find the area, first convert the value of sample mean $y>17500$ to a z-score and then obtain the area to the right of z-score from the standard normal table provided in the book.

The z-score for the value 17500is calculated as:

$\begin{array}{c}P\left(Y>17500\right)=P\left(\frac{y-\mu }{\sigma }>\frac{17500-13300}{288}\right)\\ =P\left(z>-14.58\right)\\ =1-P\left(z\le -14.58\right)\\ =1\end{array}$

The area under the curve to the right of $P\left(z\le -14.58\right)$ is obtained as 0.000 from the standard normal table.

Therefore, the obtained probability is 1.