Question

Suppose that a basketball player different from the

ones in Example 2.5-2 and in Exercise 2.5-1 can make a
free throw 60% of the time. Let X equal the minimum
number of free throws that this player must attempt to
make a total of 10 shots.
(a) Give the mean, variance, and standard deviation of X.
(b) Find P(X = 16).

Answer 1

a. Mean = 6

Variance = 2.4

Standard Deviation = 1.55

b. P(X=16) = 0.124

Step-by-step explanation:

Given

n = Total shots = 10

p = Probability of success = 60%

p = 60/100

p= 0.6

q = Probability of failure

q = 1-p

q = 1 - 0.6

q = 0.4

a.

Mean = np

Mean = 10 * 0.6

Mean = 6

Variance = npq

Variance = 10 * 0.6 * 0.4

Variance = 2.4

Standard Deviation = √Variance

Standard Deviation = √2.4

Standard Deviation = 1.549193338482966

Standard Deviation = 1.55 --------- approximated

b.

We have X = 16

x = 10

Assume that the events "success" on the various throws are independent.

The 10th success came on the 16th attempt

So, the player had exactly 10 successes and 6 failures on 16th trial

So Probability = nCr 0.6^10 * 0.4^6

Where n = 15 and r = 9 (number of attempts and success before the 16th trial)

15C9 * 0.6^10 * 0.4^6

= 5005 * 0.0060466176 * 0.004096

= 0.123958563176448

= 0.124 ------ Approximated