Any help appreciated. 1. In a student population, the height of women is normall

Question

Any help appreciated.

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 q20, 9so so that 20 percent of women are shorter than g2o 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 n = 7 is chosen at random. What is the probability that the sample mean falls in the interval [63, 64]?

Explanation / Answer

a)

probability of have height greater then 64 inches =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 percenitle ; critical value of z =-0.8416

therfore 20th percentile of height =mean +z*Std deviation=63.5-0.8416*2.9=61.0593

for 80th percenitle ; critical value of z =0.8416

therfore 80th percentile of height =mean +z*Std deviation=63.5+0.8416*2.9=65.9407

c)

probability of having height less then 72=P(X<72)=P(Z<(72-63.5)/2.9)=P(Z<2.9310)=0.9983

let sample size =n

therefore probability of having at least one women height more then 72 inches=1-P(none have height more then 72 inches) =1-(0.9983)n =0.9

(0.9983)n =0.1

taking log on both sides and solving:

n*ln(0.9983)=ln(0.1)

n=1362

d)

for sample size n=7;

std error =std deviation/(n)1/2 =2.9/(7)1/2 =1.0961

hence probability to fall 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)=0.6759-0.3241 =0.3517

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.