Q. 3.143 - Elementary Statistics

Question

Brain Weights. In $1905$ , R. Pearl published the article "Biometrical Studies on Man. I. Variation and Correlation in Brain Weight"  in which he studied the brain weights of Swedish males. The brain weights of a random sample of $225$ Swedish men have a mean of  $1.40 kg$ and standard deviation $0.11 kg$.

a. Construct a graph
b. Apply Property $1$ of the empirical rule to make pertinent statements about the observations in the sample.
c. Repeat part (b) for Property $2$ of the empirical rule.
d. Repeat part (b) for Property $3$ of the empirical rule.

Step-by-Step Solution

Answer

Part(a) Required graph is given below.

Part(b) By property $1$of empirical rule $68\%$ of sample to make statement about observation in sample.

Part(c)  By property $2$ of empirical rule $95\%$ of sample to make statement about observation in sample.

Part(d)  By property $3$ of empirical rule $99.7\%$ of sample to make statement about observation in sample.

Part(a) Step 1 : Given information

We are given that the brain weights of a random sample of $225$ Swedish men have a mean of  $1.40 kg$ and standard deviation $0.11 kg$

Part(a) Step 2 : Simplify

For drawing the graph we can first find how many men in the sample have brain weight within one, two , three standard deviation to either side of mean. 

It means we need to find $\overline{x}-3s, \overline{x}-2s, \overline{x}-s ,\overline{x}, \overline{x}+3s, \overline{x}+2s, \overline{x}+s$ .

It is already discussed in Step $4,6,8$.

Required graph is 

Part(b) Step 1 : Given information

We are given that the brain weights of a random sample of $225$ Swedish men have a mean of $1.40 kg$ and standard deviation $0.11 kg$

Part(b) Step 2 : Simplify

As we know by Property I of the empirical rule, nearly $68\%$ of the men in the sample have brain weight within one standard deviation to either side of the mean.

Now, 

$225\times \frac{68}{100}=153$

One standard deviation to either side of the mean is from
$$\begin{gathered} \overline{x}-s=1.40-0.11=1.29 \\ \overline{x}+s=1.40+0.11=1.51 \end{gathered}$$

From above calculation , we can interpret that,

Approximately $153$ of the $225$  men in the sample have brain weight between $1.29 kg$ and $1.51 kg$

Part(c) Step 1 : Given information

We are given that the brain weights of a random sample of $225$ Swedish men have a mean of $1.40 kg$ and standard deviation $0.11 kg$

Part(c) Step 2 : Simplify

As we know by Property 2 of the empirical rule, nearly$95\%$ of the men in the sample have brain weight within two standard deviations to either side of the mean.

Now,

$225\times \frac{95}{100}=213.75\approx 214$

 Two standard deviations to either side of the mean is from
$$\begin{gathered} \overline{x}-2s=1.40-2\left(0.11\right)=1.18 \\ \overline{x}+2s=1.40+2\left(0.11\right)=1.62 \end{gathered}$$

From above calculation we can say that,

Approximately $214$ of the $225$ men in the sample have brain weight between $1.18 kg$ and $1.62 kg$.

Part(d) Step 1 : Given information

We are given that the brain weights of a random sample of $225$ Swedish men have a mean of  $1.40 kg$ and standard deviation $0.11 kg$

Part(d) Step 2 : Simplify

As we know by Property 3 of the empirical rule, nearly $99.7\%$ of the men in the sample have brain weight within three standard deviations to either side of the mean.

Now, 

$225\times \frac{99.7}{100}=224.32\approx 224$

Three standard deviations to either side of the mean is from
$$\begin{gathered} \overline{x}-3s=1.40-3\left(0.11\right)=1.07 \\ \overline{x}+3s=1.40+3\left(0.11\right)=1.73 \end{gathered}$$

From above calculation , we can interpret that

Approximately $224$ of the $225$ men in the sample have brain weight between $1.07 kg$ and $1.73 kg$