(Enter your answer as a number accurate to 4 decimal places.)

Question

Based on historical data, your manager believes that 29% of the company's orders come from first-time customers. A random sample of 172 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.28 and 0.46?

 (Enter your answer as a number accurate to 4 decimal places.)

Answer 1

This one you have to leverge the rules for a sampling distribution of a proportion. That is we can model the distribution of these sample proportions with a probability model that is N(p, sqrt(pq/n))

Don't forget to check to ensure the independence, randomization, success/failure and 10% conditions are met. Proceed.

To find the probability the sample proportion of 172 orders falls between .28 and .46. We work the problem:

p = .29 = the belief/expectation

SD(phat) = sqrt(.29*.71/172) = .0346

P ( .28< phat < .46) = P(( .28 - .29)/SD(phat) < Z < (.46 -.29)/SD(phat)) = (.01/.0346 < Z < .17/.0346)

or if using a TI-84

normcdf(.28,.46, .29, .0346) = .6137