Question

Suppose a large shipment of laser printers contained 18% defectives. If a sample of size 340 is selected, what is the probability that the sample proportion will be greater than 13%? Round your answer to four decimal places.

Answer 1

Answer:

The probability that the sample proportion will be greater than 13% is 0.99693.

Step-by-step explanation:

We are given that a large shipment of laser printers contained 18% defectives. A sample of size 340 is selected.

Let $\hat p$ = the sample proportion of defectives.

The z-score probability distribution for the sample proportion is given by;

                         Z  =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} }$  ~ N(0,1)

where, p = population proportion of defective laser printers = 18%

            n = sample size = 340

Now, the probability that the sample proportion will be greater than 13% is given by = P($\hat p$ > 0.13)

         P($\hat p$ > 0.13) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} }$ > $\frac{0.13-0.18}{\sqrt{\frac{0.13(1-0.13)}{340}} }$ ) = P(Z > -2.74) = P(Z < 2.74)

                                                                      = 0.99693

The above probability is calculated by looking at the value of x = 2.74 in the  table which has an area of 0.99693.