Determining the Mean, Variance, and Standard Deviation of a Discrete Random Variable Visit the website for more services: https://cristinamontenegro92.wixsite.com/onevs

1. Mean, Variance and Standard
Deviation of a Discrete Random
Variable

2. Learning Competencies
The learner will be able to:
1. Illustrate the mean, variance and standard
deviation of a discrete random variable; and
2. Calculate the mean, variance, and standard
deviation of a discrete random variable.

3. The mean of a discrete random variable X is also called
the expected value of X. It is the weighted average of
all the values that the random variable X would assume
in the long run.
The discrete random variable X assumes values or
outcomes in every trial of an experiment with their
corresponding probabilities. The expected value of X is
the average of the outcomes that is likely to be
obtained if the trials are repeated over and over again.
The expected value of X is denoted by E(X).

4. Formula
where X=discrete random variable
x=outcome or value of the RV
P(x)=probability of the outcome x

5. Example 1 A researcher surveyed the
households in a small town. The random
variable X represents the number of college
graduates in the households. The probability
distribution of X is shown below.
Find the mean or expected value of X.
x 0 1 2
P(x) 0.25 0.50 0.25

6. Solution:
The expected value is 1. So the average number
of college graduates in the household of the
small town is one.
x 0 1 2
P(x) 0.25 0.50 0.25
x P(x) 0 0.50 0.50

7. Example 2 A random variable X has the
probability distribution. Calculate E(X).
x P(x)
1 0.10
2 0.20
3 0.45
4 0.25

8. Solution:
So E(x)=2.85
x P(x) xP(x)
1 0.10 0.10
2 0.20 0.40
3 0.45 1.35
4 0.25 1.00

9. Example 3 A security guard recorded the number of
people entering the bank every hour during one
working day. The random variable X represents the
number of people who entered the bank. The
probability distribution of X is shown below.
What is the expected number of people who enters the
bank every hour?
x P(x)
0 0
1 0.1
2 0.2
3 0.4
4 0.2
5 0.1

10. Solution:
Therefore, the average number of people entering the
bank every hour during that working day is three.
x P(x) xP(x)
0 0 0.0
1 0.1 0.1
2 0.2 0.4
3 0.4 1.2
4 0.2 0.8
5 0.1 0.5

11. The variance of a random variable X is denoted
by It can likewise be written as Var(X). The
variance of random variable is the expected
value of the square of the difference between
the assumed value of random variable and the
mean. The variance of X is:
where:
x=outcome
= population mean
P(x)=probability of the outcome

12. The larger the value of the variance, the farther are the values of
X from the mean. The variance is tricky to interpret since it
uses the square of the unit of the measure of X. So, it is easier
to interpret the value of the standard deviation because it
uses the same unit of measure of X. The standard deviation of
a discrete random variable X is written as It is the square
root of the variance. The standard deviation is computed as:

13. Example 4 Determine the variance and the
standard deviation of the following probability
mass function.
1. Find the expected value.
2. Subtract the expected value from each outcome. Square each
difference.
3. Multiply each squared difference by the corresponding probability.
4. Sum up all the figures obtained in Step3.
x P(x)
1 0.15
2 0.25
3 0.30
4 0.15
5 0.10
6 0.05

14. Solution:
x P(x) xP(x)
1 0.15
2 0.25
3 0.30
4 0.15
5 0.10
6 0.05

15. Solution:
x P(x) xP(x)
1 0.15 0.15 -1.95 3.8025 0.570375
2 0.25 0.5 -0.95 0.9025 0.225625
3 0.30 0.9 0.05 0.0025 0.00075
4 0.15 0.6 1.05 1.1025 0.165375
5 0.10 0.5 2.05 4.2025 0.42025
6 0.05 0.3 3.05 9.3025 0.465125

16. Example 5 Determine the variance and the
standard deviation of the following mass
function.
x P(x)
0 0.1
1 0.2
2 0.3
3 0.3
4 0.1

17. Solution:
x P(x) xP(x)
0 0.1
1 0.2
2 0.3
3 0.3
4 0.1

18. Solution:
x P(x) xP(x)
0 0.1 0 -2.1 4.41 0.441
1 0.2 0.2 -1.1 1.21 0.242
2 0.3 0.6 -0.1 0.01 0.003
3 0.3 0.9 0.9 0.81 0.243
4 0.1 0.4 1.9 3.61 0.361

19. Example 6 A discrete random variable X has the
probability distribution.
Find the variance and the standard deviation of
X.
x P(x)
0 0.12
1 0.25
2 0.18
3 0.35
4 0.10

20. Solution:
x P(x) xP(x)
0 0.12
1 0.25
2 0.18
3 0.35
4 0.10

21. Solution:
x P(x) xP(x)
0 0.12 0 -2.06 4.2436 0.509232
1 0.25 0.25 -1.06 1.1236 0.2809
2 0.18 0.36 -0.06 0.0036 0.000648
3 0.35 1.05 0.94 0.8836 0.30926
4 0.10 0.4 1.94 3.7636 0.37636