Question

Based on historical data, your manager believes that 33% of the company's orders come from first-time customers. A random sample of 163 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is greater than 0.28

Answer 1

Answer:

$P(x > 0.28) = 1$

Step-by-step explanation:

Given

$p = 33\%$ --- orders from first time customer

$n =163$ --- samples

Required

$P(x >0.28)$

First, calculate the mean

$\bar x = np$

$\bar x = 163 * 33\%$

$\bar x = 53.79$

Next, the standard deviation

$\sigma = \sqrt{\bar x * (1 - p)$

$\sigma = \sqrt{53.79* (1 - 33\%)$

$\sigma = \sqrt{53.79* (1 - 0.33)$

$\sigma = \sqrt{53.79* 0.67$

$\sigma = 6.00$

For $P(x >0.28)$,

The z score is:

$z = \frac{x - \bar x}{\sigma}$

$z = \frac{0.28 - 53.79}{6}$

$z = \frac{-53.51}{6}$

$z = -8.92$

So:

$P(x > 0.28) = P(z > -8.92)$

From z probability:

$P(z > -8.92) =1$

Hence:

$P(x > 0.28) = 1$