Test Unit 3 Test [Chs. 6 & 7] Time Remaining: 01 55:44 Submt Test This Question:

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Test Unit 3 Test [Chs. 6 & 7] Time Remaining: 01 55:44 Submt Test This Question: 1 pt 8 of 25 (0 complete) v This Test: 25 pts possible a Queston Help Women have head circumferences that are normally distributed with a mean given by 21.62 in.. and a standard deviation given by o w06 in Complete parts a through c below. a. If a hat company produces women's hats so that they fit head circumferences between 20.8 in and 21.8 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 the company wants to produce hats to fit all women except for those with the smalest 25% and the largest 2.5% head cicumferences, what head circumferences should be accommodated? The minimum head circumference accommodated should be in The maximum head circumference accommodated should be in. (Round to two decimal places as needed.) 208 in and in If this probability is high, does it c. If 13 women are randomly what is the probability that their mean head circumference is between 21.8 selected, women? that an order will likely fit each of 21 8 in.) so that they fit circumferences between 20.8 in. and The probability is Click to select your answer(s) a search

Explanation / Answer

a)P(20.8<X<21.8)=P((20.8-21.62)/0.6<Z<(21.8-21.62)/0.6)=P(-1.3667<Z<0.3)=0.6179-0.0859=0.5320

b)for smallest 2.5% ; z=-1.96

hence minimum circumference =21.62-1.96*0.6=20.44

maximum circumference =21.62+1.96*0.6=22.80

c) for std error =std deviation/(n)1/2 =0.166

hence P(20.8<X<21.8)=P((20.8-21.62)/0.166<Z<(21.8-21.62)/0.166)=P(-4.9276<Z<1.0817)=0.8603-0.000=0.8603

option A is correctr

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