SOLUTION: A basketball player has a free throw average of 53.2% he is about to take 5 free throws. how do i find the mean and standard deviation and the probability that he will miss 4 out o
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Question 724404: A basketball player has a free throw average of 53.2% he is about to take 5 free throws. how do i find the mean and standard deviation and the probability that he will miss 4 out of the 5
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A basketball player has a free throw average of 53.2% he is about to take 5 free throws. how do i find the mean and standard deviation and the probability that he will miss 4 out of the 5
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The mean of this binomial probability distribution is np = 5*0.532 = 2.66
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The std of this binomial probability dist
is sqrt(npq) = sqrt(2.66*0.468) = 1.1157
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Cheers,
Stan H.
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