Question

During a season a basketball player attempted 306 ​three-point baskets and made 102. He also attempted another 1354 ​two-point baskets, making 676 of these. Use these counts to determine probabilities for the following questions.

​(a) Let a random variable X denote the result of a​ two-point attempt. X is either 0 or​ 2, depending on whether the basket is made. Find the expected value and variance of X.

(b) Let a second random variable Y denote the result of a​ three-point attempt. Find the expected value and variance of Y.

(c) In a game the player attempts 6 ​three-point baskets and 16 ​two-point baskets. How many points do you expect him to​score? (Hint: Use a collection of independent and identically distributed​ (iid) random​ variables, some distributed like X and some like​ Y.)

​(d) During the season the player averaged 19.2 points from​ two- and​ three-point baskets. In the game described in part​ c, he made 33 three-pointers and 11 ​two-point baskets. Does this seem typical for his​ total?

Answer 1

a) Let a random variable X denote the result of a​ two-point attempt. He attempted 1354 ​two-point baskets and made 676 of these. So

X	0	2
P(X=x)	0.5007	0.4993

The expected value of X is :

E[X] = Σx.p(x) = 0* 0.5007 + 2*0.4993 = 0.9986

and E[X²] = Σx².p(x) = 0²* 0.5007 + 2²*0.4993 = 1.9972

The variance of X is :

V[X] = E[X²] - (E[X] )² = 1.9972 - 0.9972 = 1

b)  Let a second random variable Y denote the result of a​ three-point attempt. He attempted 306 ​three-point baskets and made 102 of these. So

X	0	3
P(Y=y)	0.6667	0.3333

The expected value of X is :

E[Y] = Σy.p(y) = 0*0.6667 + 3*0.3333 = 0.9999

and E[Y²] = Σy².p(y) = 0²*0.6667 + 3²*0.3333 =2.9997

The variance of Y is :

V[Y] = E[Y²] - (E[Y] )² =2.9997 - 0.9998 = 1.9999