Women have head circumferences that are normally distributed with a mean given b

Question

Women have head circumferences that are normally distributed with a mean given by HE23.22 in and a standard deviation given by o 1.1 in. Complete parts a through c below. a. If a hat company produces women's hats so that they fit head circumferences between 22.3 in. and 23.3 in., what is the probability that a randomly selected woman will be able to fit into one of these hats? The probability is 3274 (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 20.83 in. The maximum head circumference accommodated should be 25.61 in. (Round to two decimal places as needed.) c. If 18 women are randomly selected, what is the probability that their mean head circumference is between 22.3 in and 23.3 in.? If this probability is high, does it suggest that an order of 18 hats will very likely fit each of 18 randomly selected women? Why or why not? (Assume that the hat company produces women's hats so that they fit head circumferences between 22.3 in. and 23.3 in.) The probability is (Round to four decimal places as needed.)

Explanation / Answer

Result:

If 18 women are randomly selected, what is the probability that their mean head circumference is between 22.3 in. and 23.3 in.? If this probability is high, does it suggest that an order of 18 hats will very likely fit each of 18 randomly selected women? Why or why not?

Standard error = sd/sqrt(n) = 1.1/sqrt(18) =0.2593

Z value for 22.3, z = (22.3-23.22)/0.2593   = -3.55

Z value for 23.3, z = (23.3-23.22)/0.2593   = 0.31

P( 22.3<mean x<23.3) = P( -3.55<z<0.31)

=P( z <0.31) – P( z <-3.55)

=0.6217 -0.0002

=0.6215

The probability is 0.6215, not high, it does not suggest that an order of 18 hats will very likely fit each of 18 randomly selected women.

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