Question

A professional basketball team made 37.9% of its three-point field goals in one season. If 80 three-point field goal attempts are randomly selected from the season, what is the probability that more than 35 were made? Use a TI-83, TI-83 plus, or TI-84 calculator to find the probability. Round your answer to four decimal places.

Answer 1

Answer:

The probability that more than 35 were made is 0.14686.

Step-by-step explanation:

We are given that a professional basketball team made 37.9% of its three-point field goals in one season.

80 three-point field goal attempts are randomly selected from the season.

Let $\hat p$ = sample proportion of three-point field goals made in one season.

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 proprotion of three-point field goals = 37.9% = 0.379

            n = sample of three-point field goal attempts = 80

            $\hat p$ = sample proportion of three-point field goals = $\frac{35}{80}$ = 0.4375

Now, if 80 three-point field goal attempts are randomly selected from the season, the probability that more than 35 were made is given by = P($\hat p$ > 0.4375)

P($\hat p$ > 0.4375) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} }$ > $\frac{0.4375-0.379}{\sqrt{\frac{0.4375(1-0.4375)}{80}} }$ ) = P(Z > 1.05) = 1 - P(Z $\leq$ 1.05)

                                                                       = 1 - 0.85314 = 0.14686

The above probability is calculated by looking at the value of x = 1.05 in the z table which has an area of 0.85314.