Question

based on historical data, your manager believes that 28% of the companys orders come from first-time customers. a random sample of 131 orders will be used to estimate the proportion of first-time-customers. what is the probability that the sample proportion is greater than than 0.31?
answer = \boxed{} (enter your answer as a number accurate to 4 decimal places.)

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Explanation:

Step1: Identify distribution parameters

We have a population proportion \( p = 0.28 \), sample size \( n = 131 \). For the sampling distribution of the sample proportion \( \hat{p} \), the mean is \( \mu_{\hat{p}} = p = 0.28 \), and the standard deviation (standard error) is \( \sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}} \).

Calculate \( \sigma_{\hat{p}} \):
\( \sigma_{\hat{p}}=\sqrt{\frac{0.28\times(1 - 0.28)}{131}}=\sqrt{\frac{0.28\times0.72}{131}}=\sqrt{\frac{0.2016}{131}}\approx\sqrt{0.001539}\approx0.0392 \)

Step2: Standardize the sample proportion

We want to find \( P(\hat{p}>0.31) \). First, calculate the z - score:
\( z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}=\frac{0.31 - 0.28}{0.0392}=\frac{0.03}{0.0392}\approx0.7653 \)

Step3: Find the probability

We need to find \( P(Z > 0.7653) \). Since the total area under the standard normal curve is 1, \( P(Z > 0.7653)=1 - P(Z\leq0.7653) \).

Using a standard normal table or calculator, \( P(Z\leq0.7653)\approx0.7784 \) (using a calculator for more precision: the cumulative distribution function for \( Z = 0.7653 \) gives approximately 0.7784).

Then \( P(Z > 0.7653)=1 - 0.7784 = 0.2216 \)

Answer:

0.2216