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Get Full Access to Elementary Statistics - 12 Edition - Chapter 6.5 - Problem 13bsc
Get Full Access to Elementary Statistics - 12 Edition - Chapter 6.5 - Problem 13bsc

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# Designing Hats Women have head circumferences that are

ISBN: 9780321836960 18

## Solution for problem 13BSC Chapter 6.5

Elementary Statistics | 12th Edition

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Problem 13BSC

Problem 13BSC

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).

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 22.00 in. and 23.00 in.? If this probability is high, does it suggest that an order for 64 hats will very likely fit each of 64 randomly selected women? Why or why not?

Step-by-Step Solution:

Step 1 of 1 :

Given 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.

Here mean = 22.65 and standard deviation = 0.80

a).

From the given information the Hats by Leko company produces women’s hats so that they fit head circumferences between 21.00 in. and 25.00 in.

We consider x =21 and x=25

First we consider x=25

Now we have to find z-score.

The z-score is the value decreased by the mean, divided by the standard deviation :

z =

Substitute x, and  values.

z =

z =

z = 2.93

Therefore the value of the z-score is 2.93

Then We consider x =21

z =

Substitute x, and  values.

z =

Substitute x, and  values.

z =

z = -2.06

Therefore the value of the z-score is -2.06

Determine the corresponding probability using A-2 table.

P(-2.06 < z < 2.94) = P(z < 2.94) - P(z < -2.06)

P(-2.06 < z < 2.94) = 0.9984-0.0197

P(-2.06 < z < 2.94) = 0.9787

97.87% of women can fit into the hats.

b).

Given 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.

2.5% = 2.5/100 =0.025

Determine the z-score corresponding with an area of 0.025 in table A-2 :

The z scores for the smallest 2.5% and the largest 2.5% head circumferences are –1.96 and 1.96 respectively.

The corresponding value is the mean increased by the product of z-score and standard deviation:

We have to find  and

And

c).

If 64 women are randomly selected, the probability that their mean head circumference is between 22.00 in. and 23.00 in.

We consider n=64, x =23 and x=22

First we consider x=23

Now we have to find z-score.

The z-score is the value decreased by the mean, divided by the standard deviation :

z =

Substitute x, , and n values.

z =

z =

z = 3.5

Therefore the value of the z-score is 3.5

Then We consider x =22

z =

Substitute x, , and n values.

z =

z =

z = -6.5

Therefore the value of the z-score is -6.5

Determine the corresponding probability using A-2 table.

P(-6.5 < z < 3.5) = P(z < 3.5) - P(z < -6.5)

P(-2.06 < z < 2.94) = 0.9999-0.0001

P(-2.06 < z < 2.94) = 0.9998

99.98% between them.

No, the hats must fit individual women, not the mean from 64 women.

If all hats are made to fit head circumferences between 22 in. and 23 in., the hats will not fit about half those women.

Step 2 of 1