Question Help women have head circum erences that are normally distributed with

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Question Help women have head circum erences that are normally distributed with a mean given by 22 3, in and a standar deviation given by : 0.9 in Complete parts a through c below. a. If a hat company produces women's hats so that they fit head circumferences between 21.7 in. and 22.7 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 he company wants to produce hats to fit all wom en except for those with the smaallest 5% and the largest 1.5% hea circumferences, what he crumierences should be accommodated The minimum head circumference accommodated should be in The maximum head circumference accommodated should bein (Round to two decimal places as needed.) c. If 11 women are randomly selected, what is the probability that their mean head circumference is between 21.7 in. and 22.7 in.? If this probability is high, does it suggest that an order of 11 hats will very likely fit each of 11 randomly selected women? Why or why not? (Assume that the hat company produces women's hats so that they fit head circumferences between 21.7 in. and 22.7 in.) The probability is (Round to four decimal places as needed.) If this probability is high, does it suggest that an order of 11 hats will very likely fit each of 11 randomly selected women? Why or why not? O A. No, the hats must fit individual women, not the mean from 11 women. If all hats are made to fit head circumferences between 21.7 in. and 22.7 in., the hats won't fit about half of those women O B. Yes, the probability that an order of 1 1 hats will very likely fit each of 1 1 randomly selected women is 0.9029 C No, the hats must fit individual women, not the mean from 1 1 women all hats are made to h a r um erences between 2 n. n ni the ts won about 9 %at hos women. O D. Yes, the order of 11 hats will very likely fit each of 11 randomly selected women because both 21.7 in. and 22.7 in. lie inside the range found in part (b)

Explanation / Answer

a) P(21.7<X<22.7)=P((21.7-22.33)/0.9<Z<(22.7-22.33)/0.9)=P(-0.7<Z<0.4111)=0.6595-0.2420=0.4175

b)for 1.5 % lowest and highest ; z =-/+ 2.17

hence minimum head circumference =mean+z*std deviation =22.33-2.17*0.9 =20.38

maximum head circumference =mean+z*std deviation =22.33+2.17*0.9 =24.28

c)for std error of mean =std deviaiton/(n)1/2 =0.2714

hence

P(21.7<X<22.7)=P((21.7-22.33)/0.2714<Z<(22.7-22.33)/0.2714)=P(-2.3216<Z<1.3635)=0.9136-0.0101=0.9035

option A is correct

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