Based on historical data, your manager believes that 32% of the
company's orders come from first-time customers. A random sample of
178 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 = (Enter your answer as a number accurate to 4
decimal places.)
Solution
Given that,
_{} = [p ( 1 - p ) / n] = [(0.32 * 0.68) / 178 ] = 0.0350
P( < 0.29) =
= P[( - _{} ) / _{} < (0.29 - 0.32) / 0.0350]
= P(z < -0.8571)
= 0.1957
Probability = 0.1957
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