Question

1. Based on historical data, your manager believes that 31% of the company's orders come from first-time customers. A random sample of 146 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is greater than than 0.22?
Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.

2. 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 157000 dollars. Assume the standard deviation is 39000 dollars. Suppose you take a simple random sample of 84 graduates.
Find the probability that a single randomly selected salary is less than 160000 dollars.
Answer =
Find the probability that a sample of size n=84 n=84 is randomly selected with a mean that is less than 160000 dollars.
Answer =
Enter your answers as numbers accurate to 4 decimal places.

Answer 1

Answer 1. Based on historical data, your manager believes that 31% of the company's orders come from first-time customers. A random sample of 146 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is greater than than 0.22?

Solution:

p = 31% = 0.31

n = 146

the probability that the sample proportion is greater than than 0.22:

P( > 0.22) = P(Z > (-p)/√(p(1-p)/n)

P( > 0.22) = P(Z > (0.22 - 0.31)/√(0.31*0.69)/146)

P( > 0.22) = P(Z > -0.09 / 0.038276)

P( > 0.22) = P(Z > -2.35)

P( > 0.22) = 1 - P(Z < -2.35)

P( > 0.22) = 1 - 0.0094 (from Z table).

P( > 0.22) = 0.9906

Therefore, the probability that the sample proportion is greater than than 0.22 is 0.9906.

2. 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 157000 dollars. Assume the standard deviation is 39000 dollars. Suppose you take a simple random sample of 84 graduates.

Solution:

Mean, μ = 157000

Standard deviation, σ = 39000

n = 84

the probability that a single randomly selected salary is less than 160000 dollars:

P(X<160000) = P(Z<(160000 - 157000)/39000)

= P(Z < 0.08)

= 0.5319

Therefore, the probability that a single randomly selected salary is less than 16000 dollars is 0.5319.

the probability that a sample of size n=84 n=84 is randomly selected with a mean that is less than 160000 dollars.

P(X < 167000) = P((x-μ) /(σ /√n) < (160000 - 157000)/39000/√84)

= P(Z < 0.71)

= 0.7612

Therefore, the probability that a sample of size n=84 randomly selected with a mean that is less than 160000 dollars is 0.7612.