The probability that the number of free throws he makes exceeds 80 is approximately 0.5000.
The complete question is as below:-
A college basketball player makes 80% of his free throws. Over the course of the season, he will attempt 100 free throws. Assuming free throw attempts are independent, the probability that the number of free throws he makes exceeds 80 is approximate:____________.
A) 0.2000
B) 0.2266
C) 0.5000
D) 0.7734
Probability is defined as the ratio of the number of favourable outcomes to the total number of outcomes in other words the probability is the number that shows the happening of the event.
Given that,
A college basketball player makes 80% of his free throws.
Over the course of the season, he will attempt 100 free throws.
Assuming free throw attempts are independent.
We have to determine,
The probability that the number of free throws he makes exceeds 80 is.
According to the question,
P(Make a Throw) = 80% = 0.80
number of free throws n = 100
Binomial distribution:
Mean: n x p = 0.80 x 100 = 80
Then, The standard deviation is determined by using the formula;
[tex]\sigma[/tex] = [tex]\sqrt{np(1-P)[/tex]
= [tex]\sqrt{80\TIMES (1-0.80)[/tex]
= [tex]\sqrt16[/tex]
= 4
Therefore,
To calculate the probability that the number of free throws he makes exceeds 80 we would have to make the following calculation:
P ( x> 80) = 1 - P ( x < 80)
To calculate this value via a normal distribution approximation:
P ( [tex]Z < \dfrac{80-80}{4})[/tex] = 1-P(Z < 0 ) = 1 - 0.50 = 0.500
Hence, The probability that the number of free throws he makes exceeds 80 is approximately 0.5000.
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