13.Designing Hats. Women have head circumferences that are normally distributed
ID: 3157664 • Letter: 1
Question
13.Designing Hats. Women have head circumferences that are normally distributed with a mean of 22.65 in. And a standard deviation of 0.80 in.(based on data from the National Health and Nutrition Examination Survey) Please show all work used to derive answer. Thank You. a. If the Hats by Leko company produces women's hats so that they fit head circumferences between 21.00 in. And 25.00 in., what percentage of women can fit into these hats? 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? c. If 64 women are randomly selected, what is the probability that their mean head circumference is between 32.00 in. And 23.00 in.? If this probability is high, does it suggest that an order for 64 hats has will very likely fit each of 64 randomly selected women? Why or why not? 13.Designing Hats. Women have head circumferences that are normally distributed with a mean of 22.65 in. And a standard deviation of 0.80 in.(based on data from the National Health and Nutrition Examination Survey) Please show all work used to derive answer. Thank You. a. If the Hats by Leko company produces women's hats so that they fit head circumferences between 21.00 in. And 25.00 in., what percentage of women can fit into these hats? 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? c. If 64 women are randomly selected, what is the probability that their mean head circumference is between 32.00 in. And 23.00 in.? If this probability is high, does it suggest that an order for 64 hats has will very likely fit each of 64 randomly selected women? Why or why not? Please show all work used to derive answer. Thank You. a. If the Hats by Leko company produces women's hats so that they fit head circumferences between 21.00 in. And 25.00 in., what percentage of women can fit into these hats? 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? c. If 64 women are randomly selected, what is the probability that their mean head circumference is between 32.00 in. And 23.00 in.? If this probability is high, does it suggest that an order for 64 hats has will very likely fit each of 64 randomly selected women? Why or why not?Explanation / Answer
a)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 21
x2 = upper bound = 25
u = mean = 22.65
s = standard deviation = 0.8
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -2.0625
z2 = upper z score = (x2 - u) / s = 2.9375
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.019580079
P(z < z2) = 0.998345649
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.97876557 = 97.876557% [ANSWER]
***************************
b)
Hence, it's the middle 95% of women.
As the middle area is
Middle Area = P(x1<x<x2) = 0.95
Then the left tailed area of the left endpoint is
P(x<x1) = (1-P(x1<x<x2))/2 = 0.025
Thus, the z score corresponding to the left endpoint, by table/technology, is
z1 = -1.959963985
By symmetry,
z2 = 1.959963985
As
u = mean = 22.65
s = standard deviation = 0.8
Then
x1 = u + z1*s = 21.08202881
x2 = u + z2*s = 24.21797119
Hence, between 21.08202881 and 24.21797119 in. [ANSWER]
*******************************
c)
Hi! I am assuming you meant 22.00 to 23.00 in. In that case:
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 22
x2 = upper bound = 23
u = mean = 22.65
n = sample size = 64
s = standard deviation = 0.8
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -6.5
z2 = upper z score = (x2 - u) * sqrt(n) / s = 3.5
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 4.016E-11
P(z < z2) = 0.999767371
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.999767371 [ANSWER]
No, because this talks about the means, not the individual heads. [ANSWER]
**********************************
Hi! If c has a different given (it certainly has a typo), please resubmit this question together with the correct given. That way we can continue helping you! Thanks!
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.