Maths(四)---Probability and Statistics(6)

Expected Value

例一.

The random variable X has the distribution shown below

Solution

The expected value of X = E(X) = 1 × 0.3 + 2 × 0.2 + 3 × 0.4 + 4 × 0.1 = 2.3

例二.

The random variable X a probability distribution given by

P(X = x) =c/x ,where x = 1, 2, 3, 4, 5.

(a) Find c.

(b) Find E(X)

Solution

(a) c(1+1/2+1/3+1/4+1/5)=1

c =60/137

(b) E(X) =1× c + 2 ×c/2+3×c/3+4×c/4+5×c/5=5c=300/13

例三.

In a game, a player tosses 3 fair coins. He wins $10 if 3 heads occur, $5 if 2heads occur, $2 if only 1 head occurs and losses $15 if no heads occur. What ishis expected gain?

Solution

His expected gain =10×1/8+5×3/8+2×3/8-15×1/8=2(dollars)

Function of a Random Variable

例一.

A salesman is employed by a computer manufacturer to sell PC’s. The salary hegets in a day is calculated by the formulag(x) = 90 + 60xwhere x is the number of PC’s he sells in that day.

Assume that in each day he may sell zero to four PC’s with probabilities listed inthe table below:

Let X be the number of PC’s sold in a day.

His expected daily salary is

E[g(X)] = g(0) p(0) + g(1) p(1) + g(2) p(2) + g(3) p(3) +g(4) p(4)

= (90)(0.05) + (90+60)(0.2) + (90+120)(0.4) + (90+180)(0.2)+ (90+240)(0.15)= 222 (dollars)

E(aX + b) = aE(X) + b .

expected daily salary=

E(90+60X)= 90 + 60E(X)= 90 + 60[(0)(0.05) + (1)(0.2) + + (2)(0.4) + (3)(0.2) + (4)(0.15)]= 222 (dollars)

Variance of a Random Variable

例一.

The probability distribution of the random variable X is shown below:

Find E(X) and V(X).

Solution

E(X) =1(0.1) + 2(0.5) + 3(0.4) = 2.3

V(X) = (1− 2.3)2 (0.1) + (2 − 2.3)2 (0.5) + (3 − 2.3)2 (0.4)= 0.41

例二.

Daily sales records for a shop selling electric appliances show that it will sell zero,one, two or three air-conditioners with the probabilities:

Solution

Expected value = (0)(0.5) + (1)(0.3) + (2)(0.15) + (3)(0.05)= 0.75

Variance= (0 – 0.75)2(0.5) + (1 – 0.75)2(0.3) + (2 – 0.75)2(0.15) + (3 – 0.75)2(0.05)= 0.7875

Standard deviation = 0.7875 = 0.8874