Question

1. The National Institute for Standards and Technology (NIST) mandates that in any store no more than 2% of the items scanned through the electronic checkout scanner should have an inaccurate price. In a recent test at a Wal-Mart a sample of 529 items was randomly selected for scanning. If 2% of the items at this Wal-Mart have an inaccurate price, what is the probability that the sample proportion of items that scan innacurately is between 2.8% and 3.5%?

Answer 1

Let $X=$ the number of successes, $X\sim B(n, p).$

Given that $n=529, p=0.02.$ Then

$$np=529(0.02)=10.58>10$$

$$n(1-p)=529(1-0.02)=518.42>10$$

We can use the normal model

$$N(p, \sqrt{{p(1-p) \over n}})=N\big(0.02, \sqrt{{0.02(1-0.02) \over 529}}\big)=$$

$$=N(0.02, 0.0061)$$

to approximate the sampling distribution of $\hat{p}$.

$$P(0.028<\hat{p}<0.035)=P({0.028-0.02 \over 0.0061}<z<{0.035-0.02 \over 0.0061})\approx$$

$$\approx P(z<2.4590)-P(z<1.3115)\approx$$

$$\approx0.99303-0.90516\approx0.0879$$

The probability that the sample proportion of items that scan innacurately is between 2.8% and 3.5% is $0.0879.$