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.

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 = sample proportion of three-point field goals made in one season.

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

                           Z  =    ~ 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

            = sample proportion of three-point field goals = = 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( > 0.4375)

P( > 0.4375) = P( > ) = P(Z > 1.05) = 1 - P(Z 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.

70

step-by-step explanation:

2*5=10

4*15=60

60+10=

70

where are the images of the functions i can only see the question you typed.

step-by-step explanation: