Respuesta :
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 [tex]\hat p[/tex] = sample proportion of three-point field goals made in one season.
The z-score probability distribution for the sample proportion is given by;
Z = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} }[/tex] ~ 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
[tex]\hat p[/tex] = sample proportion of three-point field goals = [tex]\frac{35}{80}[/tex] = 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([tex]\hat p[/tex] > 0.4375)
P([tex]\hat p[/tex] > 0.4375) = P( [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} }[/tex] > [tex]\frac{0.4375-0.379}{\sqrt{\frac{0.4375(1-0.4375)}{80}} }[/tex] ) = P(Z > 1.05) = 1 - P(Z [tex]\leq[/tex] 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.
