Question

Suppose that in a day, the probability of a car agent’'s not closing any deal is 0.35. on the other hand, the probability that he/she can close one deal is 0.3; two deals, 0.25; and three deals, 0.1. find the agent’s expected number of closed deals in a day and determine the variance and standard deviation.

Answer 1

Let X - the random variable of the number of closed deals in a day.

The mean (the agent’s expected number of closed deals in a day):

$\mu=\sum x_i\cdot P(x_i)=\\ =0\cdot0.35+1\cdot0.3+2\cdot0.25+3\cdot0.1=1.1.$

The variance:

$\sigma^2=\sum(x_i-\mu)^2\cdot P(x_i),$

$X-\mu=\\ =\{ 0-1.1, 1-1.1, 2-1.1, 3-1.1\}=\\ =\{-1.1,-0.1,0.9,1.9\},$

$\sigma^2=(-1.1)^2\cdot0.35+(-0.1)^2\cdot0.3+0.9^2\cdot0.25+\\ +1.9^2\cdot0.1=0.99.$

The standard deviation:

$\sigma=\sqrt{0.99}=0.995.$