Question

Women have head circumferences that are normally distributed with a mean given by mu equals 22.16 in​., and a standard deviation given by sigma equals 0.8 in. Complete parts a through c below.

a. If a hat company produces​ women's hats so that they fit head circumferences between 21.3 in. and 22.3 ​in., what is the probability that a randomly selected woman will be able to fit into one of these​ hats? The probability is nothing. ​(Round to four decimal places as​ needed.)

b. If the company wants to produce hats to fit all women except for those with the smallest 1.25​% and the largest 1.25​% head​ circumferences, what head circumferences should be​ accommodated?

The minimum head circumference accommodated should be ____in.

The maximum head circumference accommodated should be ____in.

​(Round to two decimal places as​ needed.)

c. If 11 women are randomly​ selected, what is the probability that their mean head circumference is between 21.3 in. and 22.3 ​in.? If this probability is​ high, does it suggest that an order of 11 hats will very likely fit each of 11 randomly selected​ women? Why or why​ not? (Assume that the hat company produces​ women's hats so that they fit head circumferences between 21.3 in. and 22.3 ​in.) The probability is ____​(Round to four decimal places as​ needed.)

If this probability is​ high, does it suggest that an order of 11 hats will very likely fit each of 11 randomly selected​ women? Why or why​ not?

A.​No, the hats must fit individual​ women, not the mean from 11 women. If all hats are made to fit head circumferences between 21.3 in. and 22.3 ​in., the hats​ won't fit about half of those women.

B.​No, the hats must fit individual​ women, not the mean from 11 women. If all hats are made to fit head circumferences between 21.3 in. and 22.3 ​in., the hats​ won't fit about 28.12​% of those women.

C.​Yes, the order of 11 hats will very likely fit each of 11 randomly selected women because both 21.3 in. and 22.3 in. lie inside the range found in part​ (b).

D.​Yes, the probability that an order of 11 hats will very likely fit each of 11 randomly selected women is 0.7188.

Answer 1

a) P(21.3 < X < 22.3)

= P(-1.075 < Z < 0.175)

= P(Z < 0.175) - P(Z < -1.075)

= 0.5695 - 0.1412

= 0.4283

b) P(X < x) = 0.125

or, x = -1.15 * 0.8 + 22.16

or, x = 21.24

P(X > x) = 0.125

or, x = 1.15 * 0.8 + 22.16

or, x = 23.08

The minimum head circumference accommodated should be 21.24 in.

The maximum head circumference accommodated should be 23.08 in.

c) P(21.3 < < 22.3)

= P(Z < -3.565 < Z < 0.058)

= P(Z < 0.058) - P(Z < -3.565)

= 0.5231 - 0.0002

= 0.5229

Option - C) Yes, the order of 11 hats will very likely fit each of 11 randomly selected women because both 21.3 in. and 22.3 in. lie inside the range found in part(b).