Chapter 7.3, Problem 59E

(a)

To determine

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

Answer to Problem 59E

Normal with mean 118 and standard deviation 4.1

Explanation of Solution

Given:

  $\begin{array}{c}\mu =188\\ \sigma =41\\ n=100\end{array}$

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.

  ${\mu }_{\overline{x}}=\mu =118$

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

  ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\cdot =\frac{41}{\sqrt{100}}=\frac{41}{10}=4.1$

Therefore the sampling distribution of the sample mean is Normal with mean $118$ and standard deviation $4.1$ .

(b)

To determine

To Calculate: the probability that $\overline{x}$ estimates $\mu$ within 3 mg/dl.

Answer to Problem 59E

53.46%

Explanation of Solution

Given:

  $\begin{array}{c}\mu =188\\ \sigma =41\\ n=100\end{array}$

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

  $\begin{array}{c}z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{185-188}{41/\sqrt{100}}=-0.73\\ z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{191-188}{41/\sqrt{100}}=-0.7\end{array}$

The associating probability using the normal probability

  $P\left(Z<-0.73\right)$ is given in the row starting with $-0.7$ and in the column starting with $.03$ of the standard normal probability $P\left(Z<0.73\right)$ is given in the row starting with $0.7$ and in the column starting with $.03$ of the standard normal probability

  $\begin{array}{c}P\left(185<\overline{X}<191\right)=P\left(-0.73<Z<0.73\right)\\ =P\left(Z<0.73\right)-P\left(Z<-0.73\right)\\ =0.7673-0.2327\\ =0.5346\\ =53.46\%\end{array}$

(c)

To determine

To find: the sense is the larger sample better.

Answer to Problem 59E

97.92%

Explanation of Solution

Given:

  $\begin{array}{c}\mu =188\\ \sigma =41\\ n=1000\\ x=185\text{or 191}\end{array}$

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 sampling distribution of the sample mean $\overline{x}$ has mean $\mu$ and standard deviation $\frac{\sigma }{\sqrt{n}}\cdot$

The z-score is

  $\begin{array}{c}z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{185-188}{41/\sqrt{1000}}=-2.31\\ z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{191-188}{41/\sqrt{1000}}=2.31\end{array}$

The associating probability using the normal probability

  $P\left(Z<-2.31\right)$ is given in the row starting with $-2.3$ and in the column starting with $.01$ of the standard normal probability $P\left(Z<2.31\right)$ is given in the row starting with $2.3$ and in the column starting with $.01$ of the standard normal probability

  $\begin{array}{c}P\left(185<\overline{X}<191\right)=P\left(-2.31<Z<2.31\right)\\ =P\left(Z<2.31\right)-P\left(Z<-2.31\right)\\ =0.9806-0.0104\\ =0.9792\\ =97.92\%\end{array}$

The large sample is “better”, the reason is that the probability that the sample mean is within 3 of the population mean is higher and thus our estimates (sample means) of the population mean are more accurate.