7.67 Brain Weights. In 1905, R. Pearl published the article "Biometrical Studies on Man. 1. Variation and Correlation in Brain Weight" (Biometrika, Vol. 4, pp. 13-104). According to the study, brain weights of $S$ wedish men are normally distributed with a mean of $1.40\mathrm{kg}$ and a standard deviation of $0.11\mathrm{kg}$

a. Determine the sampling distribution of the sample mean for samples of size $3$ Interpret your answer in terms of the distribution of all possible sample mean brain weights for samples of three Swedish men.

b. Repeat part (a) for samples of size $12$

c. Construct graphs similar to those shown in Fig. $7.4$on page 304 .

d. Determine the percentage of all samples of three Swedish men that have mean brain weights within $0.1\mathrm{kg}$ of the population mean brain weight of $1.40\mathrm{kg}$. Interpret your answer in terms of sampling error.

e. Repeat part (d) for samples of size $12$

The brain weights of Swedish men follow a normal distribution with a mean of $1.40$ and a standard deviation of $0.11$

Formula used: Standard Deviation$={\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Let $x$ represent Swedish men's brain weights. Then, as follows, $x$ follows the Normal distribution: $x~N(1.40,0.11)$

With a mean of $\mu$and a standard deviation of $\sigma$the population distance is normally distributed. The sample mean sampling distribution for sample size $3$. The sampling distribution $\overline{x}$will be normal, with a mean of ${\mu }_{\overline{x}}=\mu$and a standard deviation of ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$, where $n$is the sample size.

is then determined as follows:

Here the sample size $n=3$

Mean of$\overline{x}$is${\mu }_{\overline{x}}=\mu =1.40$

Standard deviation of $\overline{x}$is,

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

$$\begin{gathered} =\frac{0.11}{\sqrt{3}} \\ =0.0635 \end{gathered}$$

As a result, the sampling distribution becomes $\overline{x}~N(1.40,0.0635)$

As a result, the brain weights in the sample exhibit a normal distribution, with a mean of $1.40\mathrm{kg}$and a standard deviation of $0.064\mathrm{kg}$

With a mean of $\mu$and a standard deviation of $\sigma$, the population distance is normally distributed. The sample mean sampling distribution for sample size $12$is then determined as follows:

The sampling distribution $\overline{x}$will be normal as well, with a mean of ${\mu }_{\lambda }=\mu$and a standard deviation of ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$, where $n$is the sample size.

Here the sample size $n=12$

Mean of $\overline{x}$is, ${\mu }_{\dot{x}}=\mu =1.40$

Standard deviation of $\overline{x}$is,

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

$$\begin{gathered} =\frac{0.11}{\sqrt{12}} \\ =0.0317 \end{gathered}$$

As a result, the sampling distribution can be calculated as $\overline{x}~N(1.40,0.0317)$

As a result, the mean brain weights in the sample have a normal distribution, with a mean of $1.40\mathrm{kg}$ and a standard deviation of $0.032\mathrm{kg}$

Using Minitab, create the graphs as follows:

• Open Minitab, go to Graph, and select Probability distribution plot from the drop-down menu.
• To continue, select View Single Plot and click Ok.
• Select the Normal distribution and fill in the Mean and Standard Deviations.
• To create the graph, click Ok.

Let $\mu$be the average Swedish man's brain weight. The sample size is $n=3$based on the information provided. The sample mean brain weights of Swedish males are thus designated by $\overline{x}$, and they are roughly distributed with ${\mu }_{\overline{x}}=\mu$and ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}=\frac{0.1}{\sqrt{3}}=0.064$

The percentage of all three Swedish males whose mean brain weights are within $0.1\mathrm{kg}$of the population mean brain weight of $1.40\mathrm{kg}$is then about equivalent to the normal curve with parameters $1.40and0.064$and sits between $\mu -0.1$and $\mu +0.1$The associated $z-$scores are then,

$$\begin{gathered} z=\frac{(\mu -0.1)-\mu }{0.064}z=\frac{(\mu +0.1)-\mu }{0.064} \\ z=\frac{0.1}{0.064}z=\frac{0.1}{0.064} \\ z=-1.56z=1.56 \end{gathered}$$

The area under the standard normal curve between $-1.56$and $1.56$is, according to Table II.

$\Phi (1.56)-\Phi (-1.56)=0.9406-0.0594=0.8812$

As a result, the mean brain weights of the three Swedish guys were within $0.1\mathrm{kg}$ of the population mean brain weight of $1.40\mathrm{kg}$ in $88.12\%$of the samples.

Let $\mu$be the average Swedish man's brain weight. The sample size is $n=12$based on the information provided. The sample mean brain weights of Swedish males are thus designated by $\overline{x}$, and they are roughly distributed with ${\mu }_{\overline{x}}=\mu$and ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}=\frac{0.1}{\sqrt{12}}=0.032$

The fraction of $12$ Swedish guys with mean brain weights within $0.1\mathrm{kg}$ of the population average of $1.40\mathrm{kg}$ is then about equivalent to the normal curve with parameters $1.40$and $0.032$and sits between $\mu -0.1$and $\mu +0.1$The associated $z-$scores are then,

$$\begin{gathered} z=\frac{\left(\mu -0.1\right)-\mu }{0.032}z=\frac{\left(\mu +0.1\right)-\mu }{0.032} \\ z=\frac{-0.1}{0.032}z=\frac{0.1}{0.032} \\ =-3.13=3.13 \end{gathered}$$

The area under the standard normal curve between $-1.56$and $1.56$is, according to Table II.

$$\begin{gathered} \Phi (3.13)-\Phi (-3.13)=0.9991-0.0009 \\ =0.9982 \end{gathered}$$

As a result, the mean brain weights of the three Swedish guys are within $0.1\mathrm{kg}$of the population mean brain weight of $1.40\mathrm{kg}$in $99.82\%$of the samples.