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 .

Answer 1

$\mu$ = 41

$\sigma$ = 35

n = 91

For sampling distribution of mean, P($\bar{x}$ < A) = P(Z < (A - $\mu_\bar{x}$ )/$\sigma_\bar{x}$)

$\mu_\bar{x}$ = $\mu$ = 41

$\sigma_\bar{x}$ =

ulo

=

35/V91

= 3.669

1) P($\bar{x}$ < 37) = P(Z < (37 - 41)/3.669)

= P(Z < -1.09)

= 0.1379

2) P(40 < $\bar{x}$ < 45) = P($\bar{x}$ < 45) - P($\bar{x}$ < 40)

= P(Z < (45 - 41)/3.669) - P(Z < (40 - 41)/3.669)

= P(Z < 1.09) - P(Z < -0.27)

= 0.8621 - 0.3936

= 0.4685

3) Let the 30th percentile be T

P($\bar{x}$ < T) = 0.30

P(Z < (T - 41)/3.669) = 0.30

Take the value of Z corresponding to 0.30 from standard normal distribution table.

(T - 41)/3.669 = -0.52

T = 39.09

4) A value is unusual if the probability of occurrence is less than 0.05

P($\bar{x}$ < 33) = P(Z < (33 - 41)/3.669)

= P(Z < -2.18)

= 0.0146

It is unusual because the probability of the sample mean being less than 33 is 0.0146

5) P(X < 33) = P(Z < (33 - $\mu$ )/$\sigma$)

= P(Z < (33 - 41)/35)

= P(Z < -0.23)

= 0.4090

It is not unusual because the probability of the sample mean being less than 33 is 0.4090