5. Assume that the mean height of American women is 63 inches with a standartd d

Question

5. Assume that the mean height of American women is 63 inches with a standartd deviation of 2.5 inches. If you chose a woman at random, what is the probability she is taller than 65 inches? What is the probability that she is between 64 and 66 inches tall? What is the 90 percentile for women's heights? What is the interquartile range for women's heights? 6, (continued from problem #5) Now assume you take a random sample of 4 women and calculate a sample mean height. What is the probability that the mean height of the women is greater than 65 inches? What is the probability that the mean height is between 64 and 66 inches tall? 6. (continued from problem #5) Now assume you take random samples of 16 women and calculate a sample mean height. What is the mean of the sampling distribution of sample means in this case? What is the standard deviation of the sampling distribution of sample means in this case? Into what interval will such a sample mean fall 95% of the time? Find the probability that your sample mean is greater than 66 inches.

Explanation / Answer

5) a) P(X>65)=1-P(X<65)=1-P(Z<(65-63)/2.5)=1-P(Z<0.8)=1-0.7881=0.2119

b)P(64<X<66)=P((64-63)/2.5<Z<(66-63)/2.5)=P(0.4<Z<1.2)=0.8849-0.6554 =0.2295

c)for 90th percentile ; z score =1.2816

therefore corresponding value =mean +z*Std deviaiton=63+1.2816*2.5=66.20

d)for interqurtile range falls between 25 and 75th percentile ; for which z score =-/+ 0.6745

1st quartile =mean +z*Std deviaiton=63-0.6745*2.5=61.3138

3rd quartile =mean +z*Std deviaiton=63+0.6745*2.5=64.6862

hence interquartile range =3rd queartile-1st quartile =64.6862-61.3138 =3.3724

6)

std error =std deviation/(n)1/2 =2.5/(4)1/2 =1.25

a)P(X>65)=1-P(X<65)=1-P(Z<(65-63)/1.25)=1-P(Z<1.6)=1-0.9452=0.0548

b)P(64<X<66)=P((64-63)/1.25<Z<(66-63)/1.25)=P(0.8<Z<2.4)=0.9918-0.7881 =0.2037

6)

a) mean of samplng distribution =63

b)std deviation of sampling distribution =std deviation/(n)1/2 =2.5/(16)1/2 =0.625

c)for 95 th percentile ; z score =-/+ 1.96

therefore 95% confidence interval =mean -/+ z*std deviation=63-/+1.96*0.625 =61.7750 to 64.2250

b) P(X>66)=1-P(X<66)=1-P(Z<(66-63)/0.625)=1-P(Z<4.800)=1-~(1) =~ 0.0000

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