Question

based on historical data, your manager believes that 29% of the companys orders come from first - time customers. a random sample of 190 orders will be used to estimate the proportion of first - time - customers. what is the probability that the sample proportion is greater than than 0.35? answer = (enter your answer as a number accurate to 4 decimal places.)

Explanation:

Step1: Calculate the mean and standard deviation of the sampling distribution of the sample proportion

The mean of the sampling distribution of the sample proportion \(\hat{p}\) is \(\mu_{\hat{p}}=p = 0.29\).
The standard deviation of the sampling distribution of the sample proportion is \(\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}\), where \(p = 0.29\) and \(n=190\).

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Step2: Standardize the sample proportion

We want to find \(P(\hat{p}>0.35)\). We use the formula \(z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}\).
Substitute \(\hat{p} = 0.35\), \(\mu_{\hat{p}}=0.29\) and \(\sigma_{\hat{p}}\approx0.0329\) into the formula:

$$ z=\frac{0.35 - 0.29}{0.0329}=\frac{0.06}{0.0329}\approx1.82 $$

Step3: Find the probability

We know that \(P(\hat{p}>0.35)=P(Z > 1.82)\).
Since \(P(Z>z)=1 - P(Z\leq z)\), and from the standard - normal table \(P(Z\leq1.82)=0.9656\).
So \(P(Z > 1.82)=1-0.9656 = 0.0344\)

Answer:

\(0.0344\)