Question

A WNba player makes 75% of her free throws. Assume that the player takes 6 free throws during the game. The distribution for the number of made free throws in 6 attempts is.

x || 0 ||    1 || 2 || 3 || 4 || 5 || 6 ||      

P(X) || 0.0002 || 0.0044 || 0.0330 || 0.1318 || 0.2966 || 0.3560 || 0.1780 ||

1. What is the probability of making at making fewer than 3 free throws.

2. What is the probability of making at least 4 of the free throws?

3. What is the mean number of free throws you expect the player to make?

4. What is the standard deviation for the number of free throws made by this player?

Answer 1

solution:

x || 0 ||    1 || 2 || 3 || 4 || 5 || 6 ||      

P(X) || 0.0002 || 0.0044 || 0.0330 || 0.1318 || 0.2966 || 0.3560 || 0.1780 ||

.1.

probability of making at making fewer than 3 free throws.

p(x<3)= p(x=2) + p(x=1) + p(x=0)

0.0330 + 0.0044 +0.0002 =0.0372

2.

p(x4)= p(x=4) + p(x=5) + p(x=6)

=0.2966+0.3560+0.1780

=0.8306

3.

μ = X * P(X) =(0.0002×0+0.0044×1+0.033×2+0.1318×3+0.2966×4+0.356×5+0.178×6) / (0.0002+0.0044+0.033+0.1318+0.2966+0.356+0.178) = 4.5002

4.

Standard deviation =

=X ² * P(X) - ²

σ = √( (0.0002×(0-4.5002)²+0.0044×(1-4.5002)²+0.033×(2-4.5002)²+0.1318×(3-4.5002)²+0.2966×(4-4.5002)²+0.356×(5-4.5002)²+0.178×(6-4.5002)²) / (0.0002+0.0044+0.033+0.1318+0.2966+0.356+0.178) ) = 1.06037