Question

A baseball player, Mickey, who bats 310 (or .310) gets an average of 3.1 hits in ten at bats. We will assume that each time Mickey bats he has a 0.31 probability of getting a hit. This means Mickeys at bats are independent from one another. If we also assume Mickey bats 5 times during a game and that x= the number of hits that Mickey gets then the following probability mass function, p(x), and cumulative distribution function F(x) are reasonable:

x	0	1	2	3	4	5
p(x)	.156	.351	.316	a	.032	.003
F(x)	.156	.507	.823	b	.997	1.00

a. Calculate the variance of 6x.
b. Assume Mickey bats exactly 5 times (a 5-at-bat-game) in each of four consecutive games. What is the probability he gets 2 hits in exactly 2 of those 4 games? (Getting exactly 2 hits will be called a 2-hit-game.)

Answer 1

from given data:

x	P(X=x)	xP(x)	x2P(x)
0	0.156	0.00000	0.00000
1	0.351	0.35100	0.35100
2	0.316	0.63200	1.26400
3	0.142	0.42600	1.27800
4	0.032	0.12800	0.51200
5	0.003	0.01500	0.07500
total		1.5520	3.4800
	E(x) =μ=	ΣxP(x) =	1.5520
	E(x2) =	Σx2P(x) =	3.4800
	Var(x)=σ2 =	E(x2)-(E(x))2=	1.0713

a) variance of 6x =Var(6x) =6²Var(x) =36*1.0713 =38.5668

b)

P(2 hits) =0.316

therefore from binomial distribution :

probability he gets 2 hits in exactly 2 of those 4 games =P(X=2)=(₄C₂)*(0.316)²(1-0.316)² =0.2803