Question

Question 4. Score: 2.2978 Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is $153,000. Assume the standard deviation is $39,000. Suppose you take a simple random sample of 49 graduates. Round all answers to four decimal places if necessary. a. What is the distribution of X? X - N( 153,00 v, 39000 v) om b. What is the distribution of T? 7 - No 39000 x, C. For a single randomly selected graduate, find the probability that her salary is between $146,643 and $152,529. d. For a simple random sample of 49 graduates, find the probability that the average salary is between $146,643 and $152,529. e. For part d), is the assumption of normal necessary? Yes No O Score: 1.143/1.143 1.143/1.143 0/1.143 0/1.143 0/1.143 0/1.143 0/1.143

Answer 1

Answer:

Given,

mean = 153000 ,

standard deviation = 39000

sample n = 49

a)

X ~ N(153000 , 39000)

b)

Xbar ~ N(153000 , 39000/sqrt(49))

c)

P(146643 < X < 152529) = P((146643 - 153000)/39000 < (x-u)/s < (152529 - 153000)/39000)

= P(- 0.163 < z < - 0.0121)

= P(z < - 0.0121) - P(z < - 0.163)

= 0.4951729 - 0.4352592 [since from z table]

= 0.0599

d)

sample n = 49

P(146643 < X < 152529) = P((146643 - 153000)/(39000/sqrt(49)) < (x-u)/(s/sqrt(n)) < (152529 - 153000)/(39000/sqrt(4)))

= P(- 1.141 < z < - 0.085)

= P(z < - 0.085) - P(z < - 1.141)

= 0.4661307 - 0.126935 [since from z table]

= 0.3392

e)

No