Question

Based on historical data, your manager believes that 29% of the company's orders come from first-time customers. A random sample of 52 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is greater than than 0.37?

Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.

Answer 1

Solution:

29% of the company's orders come from first-time customers.

Hence, population proportion of the company's orders that come from first-time customers is, P = 29/100 = 0.29.

We want to obtain the probability that sample proportion (p) is greater than 0.37.

i.e. We have to find Pr(p > 0.37).

If nP ≥ 10 and n(1 - P) ≥ 10, then sampling distribution of sample proportion follows approximately normal distribution with mean P and standard deviation .

Where, n is sample size, P is population proportion and Q = 1-P.

We have, n = 52 and P = 0.29

nP = 52 × 0.29 = 15.08 which is greater than 10.

n(1 - P) = 52 × (1 - 0.29) = 36.92 which is greater than 10.

Hence, sampling distribution of our sample proportion (p) is approximately normally distributed with mean P and standard deviation .

i.e.

And if then

We have to find Pr(p > 0.37).

We have, P = 0.29 and Q = 1 - 0.29 = 0.71 and n = 52

Using "pnorm" function of R we get, P(Z > 1.2713) = 0.1018

Hence, the probability that sample proportion is greater than 0.37 is 0.1018.

Please rate the answer. Thank you.