The mean of the number of free throws made is 79.9
The standard deviation of throws is 3.46
Karissa is a college basketball player who makes 85% of her free throws. If Karissa takes 94 free throws then,
For binomial distribution B(94,0.85)
n = no. of throws
P = probability of the success
mean = np
= 94*0.85
= 79.9
standard normal = (npq)^0.5
= (94*0.85*0.15)^0.5
= 3.46
the mean of the number of free throws made is 79.9
and standard devation of trows is 3.46
mean = np
standard deviation = (npq)^0.5
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