Question

Annual income: The mean annual income for people in a certain city (in thousands of dollars) is 41, with a standard
deviation of 35. A pollster draws a sample of 91 people to interview.

Part 1 of 5
(a) What is the probability that the sample mean income is less than 37? Round the answer to at least four decimal places.
The probability that the sample mean income is less than 37 is?

Part 2 of 5
(b) What is the probability that the sample mean income is between 40 and 45? Round the answer to at least four decimal places.
The probability that the sample mean income is between 40 and 45 is?

Part 3 of 5
(c) Find the 30th percentile of the sample mean. Round the answer to at least one decimal place.
The 30th percentile of the sample mean is?

Part 4 of 5
(d) Would it be unusual for the sample mean to be less than 33? Round the answer to at least four decimal places
It (is/isnt)unusual because the probability of the sample mean being less than 33 is?

Part 5 of 5
(e) Do you think it would be unusual for an individual to have an income of less than 33? Explain. Assume the variable is normally distributed. Round the answer to at least four decimal places.
(yes/no), because the probability that an individual has an income less than 33 is?

Answer 1

MEAN= 41 AND S.D= 35 AND STANDARD ERROR OF MEAN= 35/SQRT(91)= 35/9.54= 3.67

A]   P ( X<37 )=P ( X−μ<37−41 )=P (X−μσ<37−413.67)

Since x−μσ=Z and 37−413.67=−1.09 we have:

P (X<37)=P (Z<−1.09)

Use the standard normal table to conclude that:

P (Z<−1.09)=0.1379

B] P ( 40<X<45 )=P ( 40−41< X−μ<45−41 )=P ((40−41)/3.67<X−μ)/σ<(45−41)/3.67)

Since Z=(x−μ)/σ , (40−41)/3.67=−0.27 and (45−41)/3.67=1.09 we have:

P ( 40<X<45 )=P ( −0.27<Z<1.09 )

Use the standard normal table to conclude that:

P ( −0.27<Z<1.09 )=0.4685

C] Z VALUE FOR CORRESPONDING 0.3 IS -0.53

Z= XBAR-MEAN/S.E

-0.53*3.67= XBAR-MEAN

XBAR=39.1

D-E] P ( XBAR<33 )=P ((XBAR−μ<33−41 )=P ((Xbar−μ)/σ<(33−41)/3.67)

Since (x−μ)/σ=Z and (33−41)/3.67=−2.18 we have:

P (X<33)=P (Z<−2.18)

Use the standard normal table to conclude that:

P (Z<−2.18)=0.0146

NO IT NOT UNUSUAL.