Research reveals that foot length of women is normally distributed with mean $9.58$inches and standard deviation $0.51$inch. This distribution is useful to shoe manufactures, shoe stores, and related merchants because it permits them to make informed decisions about shoe production, inventory, and so forth. Along theses lines, the following table provides a foot-length-to-shoe-size conversion, obtained from Payless ShoeSource.

Part (a): Sketch the distribution of women's foot length.

Part (b): What percentage of women have foot lengths between $9$and $10$inches?

Part (c): What percentage of women have foot lengths that exceed $11$inches?

Part (d): Shoe manufactures suggest that if a foot length is between two sizes, wear the larger size. Referring to the preceding table, determine the percentage of women who wear size $8$shoes; size ${11}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$shoes.

Part (e): If an owner of a chain of shoe stores intends to purchase 10,000 pairs of women's shoes, roughly how many should he purchase of size $8$? of size ${11}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$? Explain your reasoning.

Consider the given question,

The mean is $9.58$inches and standard deviation is$0.51$inches.

On sketching the distribution of women's foot length,

The z-score is found using the formula $z=\frac{x-\mu }{\sigma }$.

Substitute $$\begin{gathered} x=9, \\ \mu =9.58, \\ \sigma =0.51 \end{gathered}$$,

$$\begin{gathered} z=\frac{9-9.58}{0.51} \\ =-1.14 \end{gathered}$$

Substitute $$\begin{gathered} x=10, \\ \mu =9.58, \\ \sigma =0.51 \end{gathered}$$,

$$\begin{gathered} z=\frac{10-9.58}{0.51} \\ =0.82 \end{gathered}$$

Areas under the standard normal curve, to obtain the area between the z-scores.

Area to the left to the z-score $1.14$is $0.1271$.

Area to the left to the z-score $0.82$is $0.7939$.

Thus, the area between the z-scores is given below,

Area between z-scores$=\left(Areatotheleftof0.82\right)-\left(Areatotheleftof-1.14\right)$

$$\begin{gathered} =0.7939-0.1271 \\ =0.6668 \end{gathered}$$

Thus, $66.68\%$of women foot lengths lie between$9,10$inches.

Substitute $$\begin{gathered} x=11, \\ \mu =9.58, \\ \sigma =0.51 \end{gathered}$$is given below,

role="math" localid="1652518676185" $$\begin{gathered} z=\frac{11-9.58}{0.51} \\ =2.78 \end{gathered}$$

Areas under the standard normal curve, to obtain the area between the z-scores.

Area to the left to the z-score $2.78$is $0.9973$.

Thus, the area between the z-scores is given below,

Area to right of z-score $2.78$$=\left(Areatotheleftof2.78\right)-\left(Areatotheleftof2.78\right)$

$$\begin{gathered} =1-0.9973 \\ =0.0027 \end{gathered}$$

Thus, $0.27\%$of women foot lengths exceed $11$inches.

Consider the given table,

Foot length for size $7\frac{1}{2}$is $9.5$and foot length for size $8$is $9.6875$inches.

Therefore, if a woman has foot length between $9.5,9.6875$, then she wear shoe size of $8$.

Substitute $$\begin{gathered} x=9.5, \\ \mu =9.58, \\ \sigma =0.51 \end{gathered}$$,

$$\begin{gathered} z=\frac{9.5-9.58}{0.51} \\ =-0.16 \end{gathered}$$

Substitute $$\begin{gathered} x=9.6875, \\ \mu =9.58, \\ \sigma =0.51 \end{gathered}$$,

$$\begin{gathered} z=\frac{9.6875-9.58}{0.51} \\ =0.21 \end{gathered}$$

Areas under the standard normal curve, to obtain the area between the z-scores.

Area to the left to the z-score $0.16$is $0.4364$.

Area to the left to the z-score $0.21$is $0.5832$.

Thus, the area between the z-scores is given below,

Area between z-scores$=\left(Areatotheleftof0.21\right)-\left(Areatotheleftof-0.16\right)$

$$\begin{gathered} =0.5832-0.4364 \\ =0.1468 \end{gathered}$$

Thus,$14.68\%$of women wears size$8$shoes.

Consider the given table,

Foot length for size $11$is $10.6875$and foot length for size $11\frac{1}{2}$is $10.8125$inches. Therefore, if a woman has foot length between $10.6875,10.8125$, then she wear shoe size of $11\frac{1}{2}$.

Substitute $$\begin{gathered} x=10.6875, \\ \mu =9.58, \\ \sigma =0.51 \end{gathered}$$,

$$\begin{gathered} z=\frac{10.6875-9.58}{0.51} \\ =2.17 \end{gathered}$$

Substitute $$\begin{gathered} x=10.8125, \\ \mu =9.58, \\ \sigma =0.51 \end{gathered}$$,

$$\begin{gathered} z=\frac{10.8125-9.58}{0.51} \\ =2.42 \end{gathered}$$

Areas under the standard normal curve, to obtain the area between thez-scores.

Area to the left to the z-score $2.17$is $0.985$.

Area to the left to the z-score $2.42$is $0.9922$.

Thus, the area between thez-scores is given below,

Area between z-scores$=\left(Areatotheleftof2.42\right)-\left(Areatotheleftof2.17\right)$

$$\begin{gathered} =0.9922-0.9850 \\ =0.0072 \end{gathered}$$

Thus, $0.72\%$of women wears size $11\frac{1}{2}$shoes.

It is known that the owner purchases $10,000$pairs of women shoes.

From part (d),

$$\begin{gathered} \left(Numberofshoes \\ tobepurchased\right)=10,000\times 14.68\% \\ =10,000\times 0.1468 \\ =1,468 \end{gathered}$$

Thus, the number of shoes to be purchased of size$8$is$1,468$.

It is known that the owner purchases $10,000$pairs of women shoes.

From part (d),

$$\begin{gathered} \left(Numberofshoes \\ tobepurchased\right)=10,000\times 0.72\% \\ =10,000\times 0.0072 \\ =72 \end{gathered}$$

Thus, the number of shoes to be purchased of size$11\frac{1}{2}$is$72$.