ALL WORK NEEDS TO BE SHOWN TO RECEIVE POINTS Thanks Adult US

ALL WORK NEEDS TO BE SHOWN TO RECEIVE POINTS! Thanks.

Adult US women are known to have cholesterol levels that are N(189 mg/dL, 25 mg/dL).

a.What % of US women would you expect to have cholesterol levels below 135?

b.Estimate the 25^th percentile of these cholesterol levels.

c.Estimate the interquartile range of these cholesterol levels.

d.The cholesterol levels of 149 adult female teachers at a university are measured. What is the probability that the average cholesterol level of these teachers is greater than 195 mg/dL, assuming that they are chosen randomly, are independent of each other, and are representative of all women in the US?

e.If the 149 female teachers had an average cholesterol level of 182, would you conclude that this group of women was representative of all US women? Explain your response.

Solution

(a)P(X<135) = P((X-mean)/s <(135-189)/25)

=P(Z<-2.16) =0.0154 (from standard normal table)

i.e. 1.54%

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(b)P(X<c)=0.25

--> P(Z<(c-189)/25) =0.25

--> (c-189)/25 = -0.67 (from standard normal table)

So c= 189-0.67*25=172.25

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(c)P(X<c)=0.75

--> P(Z<(c-189)/25) =0.75

--> (c-189)/25 = 0.67 (from standard normal table)

So c= 189+0.67*25=205.75

So interquartile range= 205.75-172.25=33.5

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(d)P(xbar>195) = P((xbar-mean)/(s/vn) >(195-189)/(25/sqrt(149)))

=P(Z>2.93) =0.0017 (from standard normal table)

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(e) No, becasuse it is only 149 female teachers and it is not big enough to be representative of all US women