Question

The US small business administration (SBA) provides information on the number of small business for each metropolitan area in the US. We know the population distribution is normal with a mean of 12,485 and a standard deviation of 21'937.

a) Find the probability that a random city will have more than 17000 business.

b) Find the probability that a random sample of size n=36 cities will have a mean number of small businesses greater than 17'000

c) Find the 90 percentile of sample means among those 36 cities.

d) Find the value of the 90 percentile for the population.

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	12485
std deviation   =σ=	21937.0000

probability =

P(X>17000)

=

P(Z>0.21)=

1-P(Z<0.21)=

1-0.5832=

0.4168

b)

for normal distribution z score =(X-μ)/σ
sample size       =n=	36
std error=σx̅=σ/√n=	3656.1667

probability =

P(X>17000)

=

P(Z>1.23)=

1-P(Z<1.23)=

1-0.8907=

0.1093

c)

for 90th percentile critical value of z=		1.28
therefore corresponding value=mean+z*std deviation=			17164.89

d)

for 90th percentile critical value of z=		1.28
therefore corresponding value=mean+z*std deviation=			40564.36