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.
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
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