Chapter 7.3, Problem 60E

(a)

To determine

To Explain: the sampling distribution of $\overline{x}$ .

Answer to Problem 60E

Normal with mean 9.5 and standard deviation 0.4472

Explanation of Solution

Given:

  $\begin{array}{l}\mu =9.5\\ \sigma =1.0\\ n=5\end{array}$

Formula used:

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

Calculation:

Since the population distribution is normal, the sampling distribution of the sample mean $\overline{x}$ is also normal.

The mean of the sampling distribution of the sample mean is

  ${\mu }_{\overline{x}}=\mu =9.5$

The standard deviation of the sampling distribution of the sample mean is

  ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}=\frac{1.0}{\sqrt{5}}=0.4472$

Therefore the sampling distribution of the sample mean is Normal with mean $9.5$ and standard deviation $0.4472$

(b)

To determine

To find: the probability that $\overline{x}$ estimates $\mu$ within 0.5 mm.

Answer to Problem 60E

73.72%

Explanation of Solution

Given:

  $\begin{array}{l}\mu =9.5\\ \sigma =1.0\\ n=5\\ x=9\text{ or 10}\end{array}$

Formula used:

  $z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}$

Calculation:

Since the population distribution is normal, the sampling distribution of the sample mean $\overline{x}$ is also normal.

The Z-score is

  $\begin{array}{c}z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{9-9.5}{1.0/\sqrt{5}}=-1.12\\ z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{191-188}{41/\sqrt{100}}=-0.7\end{array}$

The associating probability using the normal probability

  $P\left(Z<-1.12\right)$ is given in the row starting with $-1.1$ and in the column starting with $0.2$ of the standard normal probability $P\left(Z<1.12\right)$ is given in the row starting with $1.1$ and in the column starting with $0.2$ of the standard normal probability

  $\begin{array}{c}P\left(9<\overline{X}<10\right)=P\left(-1.12<Z<1.12\right)\\ =P\left(Z<1.12\right)-P\left(Z<-1.12\right)\\ =0.8686-0.1314\\ =0.7372\\ =73.72\%\end{array}$

(c)

To determine

To find: the probability that $\overline{x}$ falls within .5 mm of $\mu$ and the sense is the larger sample “better”.

Answer to Problem 60E

99.96%

Explanation of Solution

Given:

  $\begin{array}{l}\mu =9.5\\ \sigma =1.0\\ n=5\\ x=9\text{ or 10}\end{array}$

Formula used:

  $z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}$

Calculation:

Since the population distribution is normal, the sampling distribution of the sample mean $\overline{x}$ is also normal.

The Z-score is the

  $\begin{array}{c}z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{9-9.5}{1.0/\sqrt{50}}=-3.54\\ z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{10-9.5}{1.0/\sqrt{50}}=-3.54\end{array}$

The associating probability using the normal probability $P\left(Z<-3.54\right)$ is given in the row starting with $-3.5$ and in the column starting with $.04$ of the standard normal probability table $P\left(Z<3.54\right)$ is given in the row starting with $3.5$ and in the column starting with $.04$ of the standard normal probability

  $\begin{array}{c}P\left(9<\overline{X}<10\right)=P\left(-3.54<Z<3.54\right)\\ =P\left(Z<3.54\right)-P\left(Z<-3.54\right)\\ =0.9998-0.0002\\ =0.9996\\ =99.96\%\end{array}$

The larger sample is “ better “ because the probability that the sample mean is within $0.5$ of the population mean is higher for the larger sample and thus our estimates of the population mean are more accurate when using a larger sample.