Question

A company makes car batteries and claims 80% of its ABC batteries are good for 70 months or longer. Assume that this claim is true. Let ˆp be the proportion in a sample of 100 such ABC batteries. What is the probability that this sample proportion is within 0.05 of the population proportion.

Answer 1

The probability that this sample proportion is within 0.05 of the population proportion is 0.7888.

Step-by-step explanation:

We are given that a company makes car batteries and claims 80% of its ABC batteries are good for 70 months or longer.

A sample of 100 ABC batteries is selected.

Let $\hat p$ = sample proportion of good ABC batteries.

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

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

where, p = population proportion = 80% or 0.80

            n = sample of ABC batteries = 100

Now, the probability that the sample proportion is within 0.05 of the population proportion is given by = P(0.75 < $\hat p$ < 0.85)

As 0.80 - 0.05 = 0.75 and 0.80 + 0.05 = 0.85

    P($\hat p$ < 0.85) = P( $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ < $\frac{0.85-0.80}{\sqrt{\frac{0.80(1-0.80)}{100} } }$ ) = P(Z < 1.25) = 0.8944

    P($\hat p$ $\leq$ 0.75) = P( $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ $\leq$ $\frac{0.75-0.80}{\sqrt{\frac{0.80(1-0.80)}{100} } }$ ) = P(Z $\leq$ -1.25) = 1 - P(Z < 1.25)

                                                                   = 1 - 0.8944 = 0.1056

Therefore, P(0.75 < $\hat p$ < 0.85) = P($\hat p$ < 0.85) - P($\hat p$ $\leq$ 0.75)

                                                 = 0.8944 - 0.1056 = 0.7888.

Hence, the probability that the sample proportion is within 0.05 of the population proportion is 0.7888.