Question

Women have head circumferences that are normally distributed with a mean given by mu equals 22.14 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 21.4 in. and 22.4 ​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 1.5​% and the largest 1.5​% head​ circumferences, what head circumferences should be​ accommodated?

The minimum head circumference accommodated should be

The maximum head circumference accommodated should be . ​(Round to two decimal places as​ needed.)

c. If 19 women are randomly​ selected, what is the probability that their mean head circumference is between 21.4 in. and 22.4 ​in.? If this probability is​ high, does it suggest that an order of 19 hats will very likely fit each of 19 randomly selected​ women? Why or why​ not? (Assume that the hat company produces​ women's hats so that they fit head circumferences between 21.4 in. and 22.4 ​in.)

The probability is . ​(Round to four decimal places as​ needed.)

If this probability is​ high, does it suggest that an order of 19 hats will very likely fit each of 19 randomly selected​ women? Why or why​ not?

A. ​Yes, the order of 19 hats will very likely fit each of 19 randomly selected women because both 21.4 in. and 22.4 in. lie inside the range found in part​ (b).

B. ​No, the hats must fit individual​ women, not the mean from 19 women. If all hats are made to fit head circumferences between 21.4 in. and 22.4 ​in., the hats​ won't fit about 7.78​% of those women.

C. ​No, the hats must fit individual​ women, not the mean from 19 women. If all hats are made to fit head circumferences between 21.4 in. and 22.4 ​in., the hats​ won't fit about half of those women.

D. ​Yes, the probability that an order of 19 hats will very likely fit each of 19 randomly selected women is 0.9222.

PLEASE SHOW WORK USING A TI-84 Calculator AND By HAND

Answer 1

a)

µ =    22.14          
σ =    0.8          
we need to calculate probability for ,              
P (   21.40   < X <   22.40   )

calculator function: "normalcdf(21.40,22.40,22.14 , 0.8) "

P (   21.40   < X <   22.40   ) = 0.4499   (answer)

b)

If the company wants to produce hats to fit all women except for those with the smallest 1.5​% and the largest 1.5​% head​circumferences

The minimum head circumference accommodated should be "invnorm(0.015,22.14,0.8)" = 20.40
The maximum head circumference accommodated should be "invnorm(0.985 , 22.14 , 0.8) " = 23.88

c)

n=19

probability that their mean head circumference is between 21.4 in. and 22.4​in=0.9222

"normalcdf(21.40,22.40,22.14 , 0.8/√(19)) "

B. ​No, the hats must fit individual​ women, not the mean from 19 women. If all hats are made to fit head circumferences between 21.4 in. and 22.4 ​in., the hats​ won't fit about 7.78​% of those women.