1. In a student population, the height of women is normally distributed with mea

Question

1. In a student population, the height of women is normally distributed with mean 63.5 inches and standard deviation 2.9 inches (a) What percentage of women will have a height of greater than 64 inches? (b) Find the 20'th and 80'th percentiles of the height distribution for women. That is, find two numbers q2o, q8o so that 20 percent of women are shorter than q20 and 80 percent of women are shorter than qso- (c) We want to choose a sample of n women so that, with probability of at least 90 percent, at least one woman in the sample is more than 72 inches tall. What is the smallest value of n E N with this property? (d) Assume that a sample of size 7 is chosen at random. What is the probability that the sample mean falls in the interval [63, 64]?

Explanation / Answer

a) percentage of women have height greater then 64 =P(X>64)=1-P(X<64)=1-P(Z<(64-63.5)/2.9)

=1-P(Z<0.1724)=1-0.5684=0.4316

b) for 20th percentile ; zscore =-0.8416

therfore corresponding value=mean+z*Std deviation =63.5-0.8416*2.9=61.06

for 80th percentile ; zscore =0.8416

therfore corresponding value=mean+z*Std deviation =63.5+0.8416*2.9=65.94

c)

for probability of height more then 72 inches=P(X>72)=1-P(X<72)=1-P(Z<(72-63.5)/2.9)=1-P(Z<2.9310)

=1-0.9983=0.0017

let sample size requried =n

therfore probability that at least one women have height more then 72 =1-P(none have height more then 72)>0.9

1-(1-0.0017)n >0.9

(0.9983)n <0.1

taking log and solving

n>1361.99

n=1362

d)

for n=7 ; std error of mean =std deviaiton/(n)1/2 =2.9/(7)1/2 =1.0961

therefore probability that sample mean fals in the interval {63,64} =P(63<X<64)

=P((63-63.5)/1.0961<Z<(64.-63.5)/1.0961) =P(-0.4562<Z<0.4562)=0.6758-0.3241 =0.3517

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