Question

Based on historical data, your manager believes that 36% of the company's orders come from first-time customers. A random sample of 195 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.34 and 0.49

Answer 1

Answer:

$P(0.34 <\hat p<0.49)$

And the distribution for the sample proportion is given by;

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

And we can find the mean and deviation for the sample proportion:

[te]\mu_{\hat p}= 0.36[/tex]

$\sigma_{\hat p} =\sqrt{\frac{0.36(1-0.36)}{195}}= 0.0344$

And we can use the z score formula given by:

$z = \frac{0.34 -0.36}{0.0344}= -0.582$

$z = \frac{0.49 -0.36}{0.0344}= 3.782$

And we can use the normal distribution table and we got:

$P(-0.582 <z< 3.782) =P(z<3.782)-P(z<-0.582)=0.99992-0.2803= 0.71962$

Step-by-step explanation:

For this case we know that the sample size is n =195 and the probability of success is p=0.36.

We want to find the following probability:

$P(0.34 <\hat p<0.49)$

And the distribution for the sample proportion is given by;

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

And we can find the mean and deviation for the sample proportion:

$\mu_{\hat p}= 0.36$

$\sigma_{\hat p} =\sqrt{\frac{0.36(1-0.36)}{195}}= 0.0344$

And we can use the z score formula given by:

$z = \frac{0.34 -0.36}{0.0344}= -0.582$

$z = \frac{0.49 -0.36}{0.0344}= 3.782$

And we can use the normal distribution table and we got:

$P(-0.582 <z< 3.782) =P(z<3.782)-P(z<-0.582)=0.99992-0.2803= 0.71962$