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Answer:
The probability that the number of free throws he makes exceeds 80 is approximately 0.50
Step-by-step explanation:
According to the given data we have the following:
P(Make a Throw) = 0.80%
n=100
Binomial distribution:
mean: np = 0.80*100= 80
hence, standard deviation=√np(1-p)=√80*0.20=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)
You could calculate this value via a normal distributionapproximation:
P(Z<(80-80)/4)=1-P(Z<0)=1-50=0.50
The probability that the number of free throws he makes exceeds 80 is approximately 0.50
The probability that the number of free throws he makes exceeds 80 is approximately 0.5000.
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: [tex]n \times p = 0.80 \times 100 = 80[/tex]
Then, The standard deviation is determined by using the formula;
[tex]= \sqrt{np(1-p)} \\\\=\sqrt{80\times (1-0.80)}\\\\= \sqrt{80 \times 0.20 } \\\\= \sqrt{16} \\\\= 4[/tex]
Therefore,
To calculate the probability that the number of free throws he makes exceeds 80 we would have to make the following calculation:
[tex]P(X>80)= 1- P(X<80)[/tex]
To calculate this value via a normal distribution approximation:
[tex]P(Z<\dfrac{80-80}{4})=1-P(Z<0)=1-0.50=0.5000[/tex]
Hence, The probability that the number of free throws he makes exceeds 80 is approximately 0.5000.
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