Foot Lengths of Women Assume that foot lengths of women are normally distributed with a mean of 9.6 in. and a standard deviation of 0.5 in., based on data from the U.S. Army Anthropometry Survey (ANSUR).

a. Find the probability that a randomly selected woman has a foot length less than 10.0 in.

b. Find the probability that a randomly selected woman has a foot length between 8.0 in. and 11.0 in.

c. Find$P_{95}=10.42\text{in}$

d. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.

The lengths of foots for women is normally distributed with mean 9.6 in. and standard deviation is 0.5 in$\left(\sigma \right)$.

a) Let X be the foot length of women.

Then,

$$\begin{gathered} X~N\left(\mu ,{\sigma }^2\right) \\ ~N\left(9.6,0.5^2\right) \end{gathered}$$

The value for z-score is,

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{10-9.6}{0.5} \\ =0.8 \end{gathered}$$

The probability that a randomly selected woman has a foot length less than 10.0 in.is given by$P\left(X<10.00\right)=P\left(Z<0.8\right)$

Refer to the standard normal table for the cumulative value of 0.800 as 0.7881.Thus,$P\left(Z<0.8\right)=0.7881$ .

Therefore, the probability that a randomly selected woman has a foot length less than 10.0 in. is 0.7881.

b) The z-scores corresponding to the foots lengths are:

$$\begin{gathered} z_1=\frac{x_1-\mu }{\sigma } \\ =\frac{8.0-9.6}{0.5} \\ =-3.2 \end{gathered}$$

$$\begin{gathered} z_2=\frac{x_2-\mu }{\sigma } \\ =\frac{11.0-9.6}{0.5} \\ =2.8 \end{gathered}$$

The probability that a randomly selected woman has a foot length between 8.0 in. and 11.0 in. is given by,

$$\begin{gathered} P\left(8.0<X<11.0\right)=P\left(-3.2<Z<2.8\right) \\ =P\left(Z<2.8\right)-P\left(Z<-3.2\right)...\left(1\right) \end{gathered}$$

Refer to the standard normal table.

The cumulative area corresponding to 2.8is 0.9974.

The cumulative area corresponding to -3.2, is 0.0007.

Substitute the values in equation (1),

$$\begin{gathered} P\left(-3.2<z<2.8\right)=P\left(z<2.8\right)-P\left(z<-3.2\right) \\ =0.9974-0.0007 \\ =0.9967 \end{gathered}$$

Therefore, the probability that a randomly selected woman with foot lengths between 8.0 in. and 11.0 in. is 0.9967.

c) Let the$P_{95}$ be the 95^th percentile of foot lengths and z be the corresponding z-score.

$$\begin{gathered} P\left(X<P_{95}\right)=0.95 \\ P\left(Z<z\right)=0.95 \end{gathered}$$

Where,$z=\frac{P_{95}-\mu }{\sigma }$ .

The z-score for$P_{95}$ from the table is 1.645.

Percentile is given by,

$$\begin{gathered} z=1.645 \\ {P}_{95}=\mu +\left(1.645\times \sigma \right) \\ =9.6+\left(1.645\times 0.5\right) \\ =10.4\text{in} \end{gathered}$$

Thus, the 95^th percentile is 10.4 in.

d) Given that sample size is 25(n),

Let $\overline{X}$be the sample mean distribution for foot lengths for 25 women.

As the population is normally distributed, the sample mean distribution would be normal.

$$\begin{gathered} {\mu }_{\overline{X}}=\mu \\ =9.6\text{in} \end{gathered}$$

And

$$\begin{gathered} {\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{0.5}{\sqrt{25}} \\ =0.1\text{in} \end{gathered}$$

Thus,

$$\begin{gathered} \overline{X}~N\left({\mu }_{\overline{X}},{{\sigma }_{\overline{X}}}^2\right) \\ ~N\left(9.6,{0.1}^2\right) \end{gathered}$$

The sample mean value of 9.8 in. has corresponding z-score as,

$$\begin{gathered} z=\frac{\overline{x}-{\mu }_{\overline{X}}}{{\sigma }_{\overline{X}}} \\ =\frac{9.8-9.6}{0.1} \\ =2 \end{gathered}$$

The probability that 25 women have foot lengths with a mean greater than 9.8 in. is given by,

$$\begin{gathered} P\left(\overline{X}>9.8\right)=P\left(Z>2\right) \\ =1-P\left(Z<2\right) \\ =1-0.9772 \\ =0.0228 \end{gathered}$$

Refer to standard normal distribution to obtain the left tailed area corresponding to the value 2.00, which is 0.0228.