Question

Women have head circumferences that are normally distributed with a mean given by μ=21.88 in​., and a standard deviation given by σ =0.6 in. Complete parts a through c below.

A.) If a hat company produces​ women's hats so that they fit head circumferences between 21.6 in. and 22.6 ​in., what is the probability that a randomly selected woman will be able to fit into one of these​ hats?

The probability is=?

B.) If the company wants to produce hats to fit all women except for those with the smallest 2.5​% and the largest 2.5%head​ circumferences, what head circumferences should be​ accommodated?

The minimum head circumference accommodated should be ?in.

The maximum head circumference accommodated should be?in.

C.) If 9 women are randomly​ selected, what is the probability that their mean head circumference is between 21.6 in. and 22.6 ​in.? If this probability is​ high, does it suggest that an order of 9 hats will very likely fit each of 9 randomly selected​women? Why or why​ not? (Assume that the hat company produces​ women's hats so that they fit head circumferences between 21.6 in. and 22.6 ​in.

The probability is=?

D.)If this probability is​ high, does it suggest that an order of 9 hats will very likely fit each of 9 randomly selected​ women? Why or why​ not?

a.)​No, the hats must fit individual​ women, not the mean from 9 women. If all hats are made to fit head circumferences between

21.6 in. and 22.6 ​in., the hats​ won't fit about 8.10​% of those women.

b.).​No, the hats must fit individual​ women, not the mean from 9 women. If all hats are made to fit head circumferences between 21.6 in. and 22.6 ​in., the hats​ won't fit about half of those women.

c.).​Yes, the order of 9 hats will very likely fit each of 9 randomly selected women because both 21.6 in. and 22.6 in. lie inside the range found in part​ (b)

D.)Yes, the probability that an order of 9 hats will very likely fit each of 9 randomly selected women is 0.9190

Answer 1

A) P(21.6 < X < 22.6)

= P((21.6 - $\mu$)/$\sigma$ < (X - $\mu$)/$\sigma$ < (22.6 - $\mu$)/$\sigma$)

= P((21.6 - 21.88)/0.6 < Z < (22.6 - 21.88)/0.6)

= P(-0.47 < Z < 1.2)

= P(Z < 1.2) - P(Z < -0.47)

= 0.8849 - 0.3192

= 0.5657

B) P(X < x) = 0.025

or, P((X - $\mu$)/$\sigma$ < (x - $\mu$)/$\sigma$) = 0.025

or, P(Z < (x - 21.88)/0.6) = 0.025

or, (x - 21.88)/0.6 = -1.96

or, x = -1.96 * 0.6 + 21.88

or, x = 20.704

P(X > x) = 0.025

or, P((X - $\mu$)/$\sigma$ > (x - $\mu$)/$\sigma$) = 0.025

or, P(Z > (x - 21.88)/0.6) = 0.025

or, P(Z < (x - 21.88)/0.6) = 0.975

or, (x - 21.88)/0.6 = 1.96

or, x = 1.96 * 0.6 + 21.88

or, x = 23.056

The minimum is 20.704

The maximum is 23.056

c) P(21.6 < $\bar x$ < 22.6)

= P((21.6 - $\mu$)/($\sigma/\sqrt n$) < ($\bar x$ - $\mu$)/($\sigma/\sqrt n$) < (22.6 - $\mu$)/($\sigma/\sqrt n$)

= P((21.6 - 21.88)/(0.6/$\sqrt 9$) < Z < (22.6 - 21.88)/(0.6/$\sqrt 9$))

= P(-1.4 < Z < 3.6)

= P(Z < 3.6) - P(Z < -1.4)

= 1 - 0.0808

= 0.9192

D) Option - C) Yes, the order of 9 hats will be very likely fit each of 9 randomly selected women because both 21.6 in. and 22.6 in lie inside the range found in part(b).