A certain industrial process is brought down for recalibration whenever the quality of the items produced falls below specifications. Let $X$ represents the number of times the process is recalibrated during a week, and assume that $X$ has the following probability mass function.

$x$ 0 1 2 3 4
$P(x)$ 0.35 0.25 0.2 0.15 0.05

a. Find the mean of $X$.
b. Find the variance of $X$.
c. What is the probability that $X$ is less than 3?

Solution

Let $X$ denote number of Pepsi drinkers in a sample of six people in the city.

a. The mean of $X$ is

$$ \begin{aligned} E(X) & = \sum_x x*P(X=x)\\ & = (0*0.35)+(1*0.25)+(2*0.2)+(3*0.15)+(4*0.05)\\ & = 1.3 \end{aligned} $$

b. To find variance, we need to find $E(X^2)$.

$$ \begin{aligned} E(X^2) & = \sum_x x^2*P(X=x)\\ & = (0*0.35)+(1*0.25)+(4*0.2)+(9*0.15)+(16*0.05)\\ & = 3.2 \end{aligned} $$

The variance of $X$ is

$$ \begin{aligned} V(X) & = E(X^2) - [E(X)]^2\\ & = 3.2 - [1.3]^2\\ & = 1.51 \end{aligned} $$

c. The probability that $X$ is less than 3 is

$$ \begin{aligned} P(X < 3) &= P(X \leq 2)\\ &= P(X=0) + P(X=1) + P(X=2)\\ &= 0.35 + 0.25 + 0.2\\ &= 0.8. \end{aligned} $$

Further Reading