You may need to use the appropriate appendix table or technology to answer this question. The president of Howahkan Distributors, Inc., believes that \( 30 \% \) of the firm's orders come from first-time customers. A random sample of 100 orders will be used to estimate the proportion of first-time customers. (a) Assume that the president is correct and \( p=0.30 \). Describe the sampling distribution of \( \bar{p} \) for this study by specifying its distribution form and parameters? (Round your answer for \( \sigma_{\bar{p}} \) to four decimal places.) \( E(\bar{p})= \) (b) What is the probability that the sample proportion \( \bar{p} \) will be between 0.20 and 0.40 ? (Round your answer to four decimal places.) distribution -- Select--- \( \hat{p} \) ) appropriate in this case. (c) What is the probability that the sample proportion will be between 0.25 and 0.35 ? (Round your answer to four decimal places.)

The sampling distribution of \( \bar{p} \) is approximately normal because the sample size is large (n=100). Since we're assuming \( p = 0.30 \), the expected value of \( \bar{p} \) is \( E(\bar{p}) = p = 0.30 \). The standard deviation of the sample proportion, \( \sigma_{\bar{p}} \), can be calculated using the formula: \[ \sigma_{\bar{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.30(0.70)}{100}} = \sqrt{0.021} \approx 0.1442. \] So, \( E(\bar{p}) = 0.30 \) and \( \sigma_{\bar{p}} \approx 0.1442 \). To find the probability that \( \bar{p} \) is between 0.20 and 0.40, we will need to standardize these values using the z-score formula: \[ z = \frac{\bar{p} - p}{\sigma_{\bar{p}}}. \] Calculating for \( \bar{p} = 0.20 \): \[ z_1 = \frac{0.20 - 0.30}{0.1442} \approx -0.6931. \] Calculating for \( \bar{p} = 0.40 \): \[ z_2 = \frac{0.40 - 0.30}{0.1442} \approx 0.6931. \] Using standard normal distribution tables or technology, we can find: \[ P(-0.6931 < Z < 0.6931) \approx P(Z < 0.6931) - P(Z < -0.6931) \approx 0.7555 - 0.2445 = 0.5110. \] Thus, the probability that the sample proportion \( \bar{p} \) will be between 0.20 and 0.40 is approximately 0.5110. Now calculating for (c), we’ll find the probability that the sample proportion will be between 0.25 and 0.35. For \( \bar{p} = 0.25 \): \[ z_3 = \frac{0.25 - 0.30}{0.1442} \approx -0.3464. \] For \( \bar{p} = 0.35 \): \[ z_4 = \frac{0.35 - 0.30}{0.1442} \approx 0.3464. \] Again using the standard normal distribution tables or technology: \[ P(-0.3464 < Z < 0.3464) \approx P(Z < 0.3464) - P(Z < -0.3464) \approx 0.6368 - 0.3632 = 0.2736. \] Therefore, the probability that the sample proportion will be between 0.25 and 0.35 is approximately 0.2736.