Question

Suppose that the average annual cost of automobile insurance is $850 with an estimated population standard deviation of $245

a.)What is the probability that a simple random sample of size 150 insured automobiles will have a sample mean insurance cost greater than $950?

b.)What is the probability that a simple random sample of size 50 insured automobile will have a sample mean insurance cost within $25 of the population mean?

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	850
std deviation   =σ=	245.0000
sample size       =n=	150
std error=σx̅=σ/√n=	20.0042

probability =

P(X>950)

=

P(Z>5)=

1-P(Z<5)=

1-1=

0.0000

b)

probability =

P(825<X<875)

=

P(-0.72<Z<0.72)=

0.7642-0.2358=

0.5284