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

`z= (0.21- 0.39)/(0.0373)= -4.829`

`z= (0.32- 0.39)/(0.0373)=-1.877`

And we can find the probability with this difference:

`P(-4.829<z< -1.877) = P(Z<-1.877) -P(Z<-4.829)=0.0303- 6.86x10^(-7)=0.0303`

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{(p (1-p))/(n)} )`

And we can use the following parameters:

`\mu_(\hat p)= 0.39`

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

And we can apply the z score formula given by:

`z = (p \\mu_(\hat p))/(\sigma_(\hat p))`

And using this formula we got:

`z= (0.21- 0.39)/(0.0373)= -4.829`

`z= (0.32- 0.39)/(0.0373)=-1.877`

And we can find the probability with this difference:

`P(-4.829<z< -1.877) = P(Z<-1.877) -P(Z<-4.829)=0.0303- 6.86x10^(-7)=0.0303`