Question 124184
A basketball player makes free throw shots 80% of the time. If the player attempts 10 free throw shots,
a. What is the expected value (mean) of the number of free throw shots the player makes
Since he either makes or misses each shot this is a binomial problem.
The mean for binomial is np
Your answer is np = 20*0.8 = 16
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b. What is the standard deviation of the number of free throw shots the player makes?
The standard deviation is sqrt(npq) = sqrt(16*0.2) = 1.7888..
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c. What is the probability that the player makes at least three shots?
You have a chart and I have a TI-83 calculator.
I get  1-binomcdf(20,0.8,2) is approximately 1.
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d. What is the probability that the player makes between 7 and 9 shots, inclusively? 
I get binomcdf(20,0.8,9)-binomcdf(20,0.8,6)= 0.00056
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Cheers,
Stan H.