Unit 3 Test
(C) $46\mathrm{mins}$
18 points possible $13/18$ answered
Question 15
Based on historical data, your manager believes that $29\mathrm{\%}$ 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.)
Next Question

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 $\boxed{{\displaystyle 0.9312}}$.