Question

A certain industrial process is brought down for recalibration whenever the quality of the items falls below specifications. The random variable X represents the number of the times the process is recalibrated during a week with the following probability mass function.

X	0	1	2	3	4
P(X)	0.35	0.25	0.20	???	0.05

(a). Find the mean and variance of X

(b) Graph the PMF and CDF of X.

(c) What is the probability that X is more than one standard deviation above the mean

Answer 1

Solution:
   Given that: The random variable X represents the number of the times the process is recalibrated during a week with the following probability mass function.

X	0	1	2	3	4
P(X)	0.35	0.25	0.2	???	0.05

Probability of X=3 is unknown.

Since it is a probability distribution, we use following property of probability distribution:

Thus we get pmf:

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

Part a) Find the mean and variance of X

where

Thus we use following table:

x	P(x)	x*P(x)	x^2 *P(x)
0	0.35	0	0
1	0.25	0.25	0.25
2	0.2	0.4	0.8
3	0.15	0.45	1.35
4	0.05	0.2	0.8

Part b) Graph the PMF and CDF of X.

x	PMF P(X)	CDF F(X)
0	0.35	0.35
1	0.25	0.6
2	0.2	0.8
3	0.15	0.95
4	0.05	1

To get CDF , we add P(x) from top to bottom

PMF of X:

CDF of X:

Part c) What is the probability that X is more than one standard deviation above the mean?

P( X > Mean + 1 * SD) = ....?

Mean = E(X)=1.3

SD = square root of variance = square root of 1.51

SD = 1.2288

Thus

P( X > 1.3 + 1 * 1.2288)

=P( X > 1.3+1.2288)

= P( X > 2.5288)

=P( X =3 ) + P( X = 4)

=0.15 + 0.05

= 0.20

Thus the probability that X is more than one standard deviation above the mean is 0.20.