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
---------------
The mean of this binomial probability distribution is np = 5*0.532 = 2.66
--------------------------
The std of this binomial probability dist 
is sqrt(npq) = sqrt(2.66*0.468) = 1.1157
---------------
Cheers,
Stan H.