Based on historical data, your manager believes that 27% of the company's orders come from first-time customers. A random sample of 65 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is greater than than 0.19? 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.)
p = 0.27, p' = 0.19, n = 65
z = (p' - p)/SE where SE = √(p(1 - p)/n)
SE = √(0.27 (1 - 0.27)/65) = 0.0551
z = (0.19 - 0.27)/0.0551 = -1.4528
P(p' > 0.19) = P(z > -1.4528) = 0.9269
Based on historical data, your manager believes that 27% of the company's orders come from first-time...
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