Question

Women have head circumferences that are normally distributed with a mean given by mu equals 21.05 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 20.5 in. and 21.5 ​in., what is the probability that a randomly selected woman will be able to fit into one of these​ hats?

The probability is ___. ​(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 3.75% and the largest 3.75% head​ circumferences, what head circumferences should be​accommodated? The minimum head circumference accommodated should be nothing in. The maximum head circumference accommodated should be nothing in. ​(Round to two decimal places as​ needed.)

c. If 7 women are randomly​ selected, what is the probability that their mean head circumference is between 20.5 in. and 21.5 ​in.? If this probability is​ high, does it suggest that an order of 7 hats will very likely fit each of 7 randomly selected​ women? Why or why​ not? (Assume that the hat company produces​ women's hats so that they fit head circumferences between 20.5 in. and 21.5 ​in.)

The probability is __. ​(Round to four decimal places as​ needed.) If this probability is​ high, does it suggest that an order of 7 hats will very likely fit each of 7 randomly selected​ women? Why or why​ not?

A.​Yes, the probability that an order of 7 hats will very likely fit each of 7 randomly selected women is 0.8975.

B.​No, the hats must fit individual​ women, not the mean from 7 women. If all hats are made to fit head circumferences between 20.5 in. and 21.5 ​in., the hats​ won't fit about half of those women.

C.​No, the hats must fit individual​ women, not the mean from 7 women. If all hats are made to fit head circumferences between 20.5 in. and 21.5 ​in., the hats​ won't fit about 10.25​% of those women.

D.​Yes, the order of 7 hats will very likely fit each of 7 randomly selected women because both 20.5 in. and 21.5 in. lie inside the range found in part​ (b).

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	21.05
std deviation   =σ=	0.8000

probability =

P(20.5<X<21.5)

=

P(-0.69<Z<0.56)=

0.7123-0.2451=

0.4672

b)

for 3.75% at both end critical values z=-/+ 1.78

minimum head circumference accommodated should be =19.63

maximum head circumference accommodated should be =22.47

c)

probability =

P(20.5<X<21.5)

=

P(-1.82<Z<1.49)=

0.9319-0.0344=

0.8975

B.​No, the hats must fit individual​ women, not the mean from 7 women. If all hats are made to fit head circumferences between 20.5 in. and 21.5 ​in., the hats​ won't fit about half of those women