Question

A bottle maker believes that 14% of his bottles are defective. If the bottle maker is accurate, what is the probability that the proportion of defective bottles in a sample of 622 bottles would be less than 11%

Answer 1

Answer:

$z = \frac{0.11-0.14}{0.0139} = -2.156$

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

$P(Z$

Step-by-step explanation:

For this case we know the following info given:

$p =0.14$ represent the population proportion

$n = 622$ represent the sample size selected

We want to find the following proportion:

$P(\hat p$

For this case we can use the normal approximation since we have the following conditions:

i) np = 622*0.14 = 87.08>10

ii) n(1-p) = 622*(1-0.14) =534.92>10

The distribution for the sample proportion would be given by:

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

The mean is given by:

$\mu_{\hat p}= 0.14$

And the deviation:

$\sigma_{\hat p}= \sqrt{\frac{0.14*(1-0.14)}{622}}= 0.0139$

We can use the z score formula given by:

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

And replacing we got:

$z = \frac{0.11-0.14}{0.0139} = -2.156$

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

$P(Z$