Question

A basketball player has made 6060​% of his foul shots during the season. Assume the shots are independent. ​

Answer 1

Answer:

a) 2.5 shots

b) 59.4 shots

c) 4.87 shots

Step-by-step explanation:

Probability of making the shot = 0.6

Probability of missing the shot = 0.4

a) The expected number of shots until the player misses is given by:

$E(X) = \frac{1}{P(X)}=\frac{1}{0.4} \\E(X) = 2.5\ shots$

The expected number of shots until the first miss is 2.5

b) The expected number of shots made in 99 attempts is:

$E=99*0.6\\E=59.4\ shots$

He is expected to make 59.4 shots

c) Let "p" be the proportion shots that the player make, the standard deviation for n = 99 shots is:

$s=\sqrt{n*p*(1-p)}\\ s= 4.87\ shots$

The standard deviation is 4.87 shots.