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.)
Solución de inteligencia artificial de Upstudy
Responder
Solución

¡Inicia sesión para desbloquear respuestas gratis!
Una plataforma de aprendizaje en la que confían millones de estudiantes y profesores reales.

Respondido por UpStudy AI y revisado por un tutor profesional
The Deep Dive
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.
