Explain C) only please, thank you. Homework: Section 6.5 Save Score: 0.4 of 1 pt

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Explain C) only please, thank you.

Homework: Section 6.5 Save Score: 0.4 of 1 pt | 8 of 13 (8 complete) HW Score: 60%, 8.4 of 14 pts 6.5.13-T Assigned Media | Question Help Women have head circumferences that are normally distributed with a mean given by -24.42 in., and a standard deviation given by: 1.1 in. Complete parts a through c below able to fit into one of these hats? The probability is 0.3499 (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 22.03 in. The maximum head circumference accommodated should be 26.81 in (Round to two decimal places as needed.) c. If 13 women are randomly selected, what is the probability that their mean head circumference is between 24.0 in. and 25.0 in.? If this probability is high, does it suggest that an order of 13 hats will very likely fit each of 13 randomly selected women? Why or why not? (Assume that the hat company produces women's hats so that they fit head circumferences between 24.0 in. and 25.0 in.) The probability is (Round to four decimal places as needed.) Enter your answer in the answer box and then click Check Answer 2 part remaining Clear All Check Answer

Explanation / Answer

c)here std error of mean =std deviation/(n)1/2 =1.1/(13)1/2 =0.3051

therefore probabilty =P(24<X<25)=P((24-24.42)/0.3051<Z<(25-24.42)/0.3051)

            =P(-1.3767<Z<1.9011)=0.9714-0.0843 =0.8870

No; as above probabilty is for mean of 13 randomely selected women not for an individual women.

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