Question

Based on historical data, your manager believes that 28% of the company's orders come from first-time...

Based on historical data, your manager believes that 28% of the company's orders come from first-time customers. A random sample of 157 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.27 and 0.47?

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

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Answer #1

Solution:

p = 28% = 0.28 , n = 157

The sampling distribution of sample propotion is approximately normal with

mean = p = 0.28 , standard deviation = √p(1-p)/n = √(0.28*0.72)/157 = 0.0358

z = (p^-mean)/standard deviation

P(0.27 < p^ < 0.47)

= P(p^<0.47)-P(p^<0.27)

= P(z<5.30)-P(-0.28)

= 1-0.3897

= 0.6103

Answer = 0.6103

Z Table Z: Areas under the standard normal curve (negative Z) Second decimal place in z 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.

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