In this arithmetic sequence, what is the common difference? -19, -11, -3, 5
1 answer:
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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 = <em>sample proportion of three-point field goals made in one season.</em>
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 = <u>0.14686</u>
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.