Chapter 10.2, Problem 10.36E

(a)

To determine

To Explain: that expect the mean of the distribution of mean’s from all possible samples to be.

Answer to Problem 10.36E

0.5, Yes

Explanation of Solution

Given:

  $\mu =0.5$

Formula used:

  $\mu =\frac{{\displaystyle \sum x}}{n}$

Calculation:

  ${\mu }_{\overline{x}}=\mu =0.5$

Therefore would the mean expected to be 0.5

The mean is the addition of all values divided by the number of certain values:

  $\begin{array}{c}\overline{x}=\frac{0.532+0.481+\mathrm{....}+0.480+0.479+0.499}{50}\\ =0.4958\end{array}$

It is observed that the mean of the 50 sample means of 0.4958 is close to the population mean of 0.5.

(b)

To determine

To Calculate: the standard deviation.

Answer to Problem 10.36E

0.325

Explanation of Solution

Given:

  $n=100$

Formula used:

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

Calculation:

Standard deviation

  $\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{\sigma }{\sqrt{100}}\\ =\frac{\sigma }{10}\end{array}$

  $\begin{array}{c}{s}_{\overline{x}}=\sqrt{\frac{{\left(0.532-0.4958\right)}^2+\mathrm{....}{\left(0.499-0.4958\right)}^2}{50-1}}\\ =0.0325\end{array}$

Since ${\sigma }_{\overline{x}}=\frac{\sigma }{10}$ is equivalent with $\sigma =10{\sigma }_{\overline{x}}$

  $s=10s_{\overline{x}}=10\left(0.325\right)=0.325$

Therefore it is estimated that the standard deviation as 0.325.