Chapter 7.3, Problem 49E

To determine

Tocalculate:The mean and the standard deviation of the sampling distribution of $\overline{x}$ .

Answer to Problem 49E

The required mean and standard deviation are 225 seconds and 18.9737 seconds respectively.

Explanation of Solution

Given information:

Number of songs in the iPod = 10,000

Mean = 225 seconds

Standard deviation = 60 seconds

Number of songs in SRS = 10

Formula used:

The mean of the sampling distribution of $\widehat{p}$is, ${\mu }_{\widehat{p}}=p$ .

The standard deviation of a sample distribution of the sample mean $\overline{x}$if the population standard deviation is $\sigma$can be written as,

  ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Calculation:

Here, given data can be written as:

Mean, $\mu =225$

Population standard deviation, $\sigma =60$

Sample size, n = 10

Know that the mean of the sampling distribution of $\widehat{p}$is, ${\mu }_{\widehat{p}}=p$ .

So, the mean of the sampling distribution of the sample mean $\overline{x}$is,

  $\begin{array}{l}{\mu }_{\overline{x}}=\mu \\ {\mu }_{\overline{x}}=225\end{array}$

Also, the standard deviation of a sample distribution of the sample mean $\overline{x}$if the population standard deviation is $\sigma$can be written as,

  ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Substitute 60 for $\sigma$and 10 for n in the above formula and simplify.

  $\begin{array}{c}{\sigma }_{\overline{x}}=\frac{60}{\sqrt{10}}\\ \approx 18.9737\end{array}$

Hence, the required mean and standard deviation are 225 seconds and 18.9737 seconds respectively.