Random Variable - Statistics

A mail order computer business has six telephone lines. Let X denotes the number of lines in use at a given specified time. Suppose that the probability density (mass) function of X is given in the following table:

...X.........0........ 1....... 2............3......... 4 .......5..........6........Otherwise....
...P(X)...0.10... 0.15.....0.20 .....0.25.... 0.20....0.06.....0.04.........0.........鈥?br>
i. What is the probability that at most 4 lines are in use?

ii. What is the probability that less than 4 lines are in use?

iii. What is the probability that between 3 and 5 lines are in use? (3 and 5 are included)

iv. What is the probability that between 2 and 4 lines are not in use? (2 and 4 are included)

v. What is the expected value of X?

vi. What is the standard deviation of X?

i) P(x 鈮?)=P(x=0)+P(x=1)+P(x=2)+(x=3)+P(x=4)

=0.1+0.15+0.2+.25+0.2= 0.9

ii) P(x<4)=P(x=0)+P(x=1)+P(x=2)+(x=3)

=0.1+0.15+0.2+.25= 0.7

iii) P(3鈮鈮?)=P(x=3)+P(x=4)+P(x=5)

=0.25+0.2+0.06=0.51

iv) P(x=0)+P(x=1)+P(x=5)+P(x=6)

=0.1+0.15+0.06+0.04=0.35

v)E(x)=饾毢xP(x)

=(0)(0.1)+(1)(0.15)+(2)(0.2)+(3)(0.25)
+(4)(0.2)+(5)(0.06)+(6)(0.04) =2.64

vi) You need to find the variance first add x^2 to the table

...X^2.........0........ 1....... 4............9......... 16 ......25.......36........Otherwise....
.X^2P(X)...0... ..0.15.....0.80 .....2.25.... ..3.2.....1.5....1.44.........0.........鈥?br>
Var=饾毢x^2P(x) - [E(x)]^2

=9.34 - (2.64)^2= 2.37

SD=sqrt(var)=sqrt(2.37)=1.54