This is the third of three statistics problems on bonus free throws. For all three problems go to the link below.
http://fractiontips.blogspot.com/2015/04/thestatisticsofbonusfreethrows.html
Question: What is the expected number and variance of free throw makes for a person with 5 bonus free throw opportunities and for a person with 10 regular (not in bonus) free throw opportunities? Provide calculations for a 60 percent free throw shooter and a 90 percent free throw shooter?
http://fractiontips.blogspot.com/2015/04/thestatisticsofbonusfreethrows.html
Question: What is the expected number and variance of free throw makes for a person with 5 bonus free throw opportunities and for a person with 10 regular (not in bonus) free throw opportunities? Provide calculations for a 60 percent free throw shooter and a 90 percent free throw shooter?
Answer: A previous post found the probability
distribution function for a person who had one bonus free throw situation.
I used the PDF to calculate expected makes and variance of
makes at 0.6 and 0.8.
The numbers in the chart for p=0.6 and p=0.8 are repeated
below for your convenience.
Calculation For
Bonus Situation


# Makes

Prob (Make=k)

p=0.6

p=0.8

0

(1p)

0.4

0.2

1

(1p) x p

0.24

0.16

2

p x p

0.36

0.64

E(Makes)

0.96

1.44


E(Makes^2)

1.68

2.72


Var(Makes)

0.7584

0.6464

The outcomes from the bonus free throw situations are
independent and identically distributed.
This means the expected outcome of the sum of n situations
is simply the sum of the expected outcome of all situations.
This also means that the variance of the sum of n situations
is simply the sum of the variance of each situation.
The sum of n identical items is the same as n multiplied by
one of the items.
Using these insights I create answers presented in chart
below.
Expected Makes and
Variance of Makes for 5
or 10 Bonus Situations


Five Bonus Situations


Prob makes shot

0.6

0.8

Expected Make Per Bonus
Situation

0.96

1.44

Variance One Bonus Situation

0.75

0.6464

Expected Number Makes

4.8

7.2

Variance Makes

3.75

3.232

Ten Bonus Situations


Prob makes shot

0.6

0.8

Expected Make Per Bonus
Situation

0.96

1.44

Variance One Bonus Situation

0.75

0.6464

Expected Number Makes

9.6

14.4

Variance Makes

7.5

6.464

Concluding Thoughts
The expected number of points from a bonus free throw
situation is 1.5 times higher for a 80 percent free throw shooter than a 60 percent
free throw shooter.
The variance of number of makes is around 14 percent lower
for the 80 percent shooter than the 60 percent shooter.
Note from previous post the probability distributions
function for the bonus free throw situation has a small right tail when the
number of bonus opportunities rises.
The probability of making 10 shots on 5 bonus opportunities is p^{10}. The high variability for p=0.6 means that a
lot of misses are likely to occur.
A useful exercise for a motivated student might be to
calculate the expected value and variance of points from a bonus situation as a
function in terms of p the probability of making one shot and then solving for
expected points and variance of points for n bonus opportunities.
People who liked this
post may also like my book on statistics and baseball.
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