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. If p is the probability mass function for x, then evaluate a?
b. If F is the cumulative distribution function of x, then what is the value of b?
c. What is the expected value of x?
d. Calculate the standard deviation of x.
e. Calculate the variance of 6x.
f. 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

Computational Table:

X	P(X)	F(X)	X*P(X)	X^2*P(X)
0	0.156	0.156	0	0
1	0.351	0.507	0.351	0.351
2	0.316	0.823	0.632	1.264
3	0.142	0.965	0.426	1.278
4	0.032	0.997	0.128	0.512
5	0.003	1.000	0.015	0.075
Total			1.552	3.48

a)

We, Know that

Sum of all probabilities = 1

Therefore, 0.156 + 0.351 + 0.316 + a + .032 + 0.03 = 1

0.858 + a = 1

a = 1 - 0.858 = 0.142

b)

And, b = a + 0.823 = 0.142 + 0.823 = 0.965

C)

d)

Standard deviation (SD):

Variance (X):

V(X) = E(X²) - [E(x)]²

V(X) = 3.48 - (1.552²)

V(X) = 1.07

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SD= V(X) = E(X2) – [E(2)]2 = V1.07 = 1.04