Question

Suppose the weight of males ages 20-39 are normally distributed with a mean 196.9 lbs. and a standard deviation 25 lbs.

A. What weight is considered the 20th percentile?

B. What is the percentile of someone who weighs 165 lbs.?

C. What is the proportion of males who weigh between 200 and 225 lbs.?

D. Suppose 45 males were selected at random and weighed, what is the proportion of males above 205?

Answer 1

A)

for 20th percentile critical value of z=		-0.84
therefore corresponding value=mean+z*std deviation=			175.9

B)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	196.9
std deviation   =σ=	25.0000

probability =

P(X<165)

=

P(Z<-1.28)=

0.1003~ 10.03 percentile

C)

probability =

P(200<X<225)

=

P(0.12<Z<1.12)=

0.8686-0.5478=

0.3208

D)

sample size       =n=	45
std error=σx̅=σ/√n=	3.7268

proportion of males above 205 :

probability =

P(X>205)

=

P(Z>2.17)=

1-P(Z<2.17)=

1-0.9850=

0.0150