Unit 3 Test
(C) $46 \mathrm{mins}$
18 points possible $13 / 18$ answered
Question 15
Based on historical data, your manager believes that $29 \%$ of the company's orders come from first-time customers. A random sample of 71 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.21 and 0.49 ?
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.)
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Step 1 ：Given that the number of orders, \(n = 71\), and the probability of success, \(p = 0.29\).

Step 2 ：Calculate the mean of the binomial distribution, which is \(np = 71 \times 0.29 = 20.59\).

Step 3 ：Calculate the standard deviation of the binomial distribution, which is \(\sqrt{np(1-p)} = \sqrt{71 \times 0.29 \times (1 - 0.29)} = 3.823467013065498\).

Step 4 ：Standardize the sample proportions 0.21 and 0.49 to z-scores using the formula \(z = \frac{x - \text{mean}}{\text{standard deviation}}\).

Step 5 ：For \(x = 0.21\), the z-score is \(z1 = \frac{0.21 - 20.59}{3.823467013065498} = -1.4855627054164149\).

Step 6 ：For \(x = 0.49\), the z-score is \(z2 = \frac{0.49 - 20.59}{3.823467013065498} = 3.7139067635410368\).

Step 7 ：Find the probability that the z-score is between the two calculated z-scores using the standard normal distribution. The probability is approximately 0.9312005450671094.

Step 8 ：Final Answer: The probability that the sample proportion is between 0.21 and 0.49 is \(\boxed{0.9312}\).