91Ó°ÊÓ

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

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

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

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,

$225×\frac{68}{100}=153$

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 $1.29kg$and $1.51kg$

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

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,

$225×\frac{95}{100}=213.75≈214$

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,

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

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,

$225×\frac{99.7}{100}=224.32≈224$

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