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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.

User Andreis
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Answer:

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

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{(\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{(\hat p(1-\hat p))/(n)} } >
\frac{0.13-0.18}{\sqrt{(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.

User Teshguru
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