Question

Based on historical data, your manager believes that 39% of the company's orders come from first-time customers. A random sample of 171 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.21 and 0.32

Answer 1

Answer:

$z= \frac{0.21- 0.39}{0.0373}= -4.829$

$z= \frac{0.32- 0.39}{0.0373}=-1.877$

And we can find the probability with this difference:

$P(-4.829$

Step-by-step explanation:

For this case we have the following info given:

$n = 171$ represent the sample size

$p =0.39$ the proportion of interest

We want to find the following probability:

$P( 0.21 < \hat p < 0.32)$

We can use the normal approximation for this case since np >10 and n (1-p) >10

For this case we know that the distribution for the sample proportion is given by:

$\hat p \sim N( p , \sqrt{\frac{p (1-p)}{n}} )$

And we can use the following parameters:

$\mu_{\hat p}= 0.39$

$\sigma_{\hat p} =\sqrt{\frac{0.39*(1-0.39)}{171}}= 0.0373$

And we can apply the z score formula given by:

$z = \frac{p \\mu_{\hat p}}{\sigma_{\hat p}}$

And using this formula we got:

$z= \frac{0.21- 0.39}{0.0373}= -4.829$

$z= \frac{0.32- 0.39}{0.0373}=-1.877$

And we can find the probability with this difference:

$P(-4.829$