Question

Based on historical data, your manager believes that 27% of the company's orders come from first-time customers. A random sample of 68 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is less than 0.29?

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

Answer 1

Solution

=  [p ( 1 - p ) / n] =   [(0.27 * 0.73) / 68] = 0.05384

P( < 0.29) =

= P[( - ) / < (0.29 - 0.27) / 0.05384]

= P(z < 0.3715)

= .6449

Proportion = 0.6449