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

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

And the distribution for the sample proportion is given by;

`\hat p \sim N(p, \sqrt{(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{(0.36(1-0.36))/(195)}= 0.0344`

And we can use the z score formula given by:

`z = (0.34 -0.36)/(0.0344)= -0.582`

`z = (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`

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{(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{(0.36(1-0.36))/(195)}= 0.0344`

And we can use the z score formula given by:

`z = (0.34 -0.36)/(0.0344)= -0.582`

`z = (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`