Question

Problem 4. The lifetime of a certain battery follows a normal distribution with a mean of 276 and standard deviation of 20 minutes. (a) What proportion of the batteries have a lifetime more than 270 minutes? (b) Find the 90th percentile of the lifetime of these batteries. (c) We took a random sample of 100 batteries. What is the probability that the sample mean of lifetimes will be less than 270 minutes?

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	276
std deviation   =σ=	20.000

probability =P(X>270)=P(Z>(270-276)/20)=P(Z>-0.3)=1-P(Z<-0.3)=1-0.3821=0.6179

b)

for 90th percentile critical value of z=		1.28
therefore corresponding value=mean+z*std deviation=			301.6

c)

sample size       =n=	100
std error=σx̅=σ/√n=	2.000

probability =P(X<270)=(Z<(270-276)/2)=P(Z<-3)=0.0013