Question

Suppose a large shipment of microwave ovens contained 12% defectives. If a sample of size 474 is selected, what is the probability that the sample proportion will be greater than 14%

Answer 1

Answer:

$P(\hat p>0.14)$

And using the z score given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

Where:

$\mu_{\hat p} = 0.12$

$\sigma_{\hat p}= \sqrt{\frac{0.12*(1-0.12)}{474}}= 0.0149$

If we find the z score for $\hat p =0.14$ we got:

$z = \frac{0.14-0.12}{0.0149}= 1.340$

So we want to find this probability:

$P(z>1.340)$

And using the complement rule and the normal standard distribution and excel we got:

$P(Z>1.340) = 1-P(Z$

Step-by-step explanation:

For this case we have the proportion of interest given $p =0.12$. And we have a sample size selected n = 474

The distribution of $\hat p$ is given by:

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

We want to find this probability:

$P(\hat p>0.14)$

And using the z score given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

Where:

$\mu_{\hat p} = 0.12$

$\sigma_{\hat p}= \sqrt{\frac{0.12*(1-0.12)}{474}}= 0.0149$

If we find the z score for $\hat p =0.14$ we got:

$z = \frac{0.14-0.12}{0.0149}= 1.340$

So we want to find this probability:

$P(z>1.340)$

And using the complement rule and the normal standard distribution and excel we got:

$P(Z>1.340) = 1-P(Z$