Answer to Question #304862 in Statistics and Probability for Ash

Question

The random variable X represents the number of taxi cabs a doctor meets on his way to the hospital. The probability distribution of X is given in the following distribution table.

X P(x)

0 0.01

1 0.15

2 0.20

3 0.25

4 0.30

5 0.09

Find the expected value and variance of X.

Answer 1

$$mean=E(X)=\sum_ix_ip(x_i)$$

$$=0(0.01)+1(0.15)+2(0.20)+3(0.25)$$

$$+4(0.30)+5(0.09)=2.95$$

$$E(X^2)=0^2(0.01)+1^2(0.15)+2^2(0.20)$$

$$+3^2(0.25)+4^2(0.30)+5^2(0.09)=10.25$$

$$Var(X)=\sigma^2=E(X^2)-(E(X))^2$$

$$=10.25-(2.95)^2=1.5475$$