This document discusses key concepts about finding the mean and variance of discrete probability distributions. It defines that the mean of a random variable X is equal to the sum of the products of each possible value of X and its corresponding probability. The variance is the weighted average of the squared differences between each value and the mean. Several examples are provided to demonstrate calculating the mean, variance, and standard deviation of distributions.

1. The session
shall begin
shortly…

3. Mean and Variance of
Discrete Probability
Distributions
Statistics & Probability
Grade 11

4. Mean and Variance of Random Variables
For a discrete random variable X
with probability distribution
function P(X), the EXPECTED VALUE
or MEAN of X is given by
      E X x P x
Mean of a Random Variable
The mean is equal to the SUM of the PRODUCTS of the
values of X and their corresponding probabilities P(X)

5. Mean and Variance of Random Variables
Remarks:
1. “The mean of the random
variable” can also mean
“the mean of the
probability distribution.”
Mean of a Random Variable

6. Mean and Variance of Random Variables
Remarks:
2. The mean of a random variable
is the weighted AVERAGE value
of all the possible values of X,
with the weights being the
probabilities of the respective
values of X.
Mean of a Random Variable

7. Find the
mean of the
given
probability
distribution.
Mean and Variance of Random Variables
Mean of a Random Variable
X P(X)
0 0.05
1 0.15
2 0.35
3 0.45

8. number
of head
X
0 1 2 3
P(X) 1/8 3/8 3/8 1/8
If three coins
are tossed,
find the
mean of the
number of
heads X that
occur.
Mean and Variance of Random Variables
Mean of a Random Variable
Solution
Recall that the pdf for the
number of heads is as
follows:

9. Mean and Variance of Random Variables
For a discrete random
variable X with probability
distribution function P(X), the
VARIANCE of X is given by
      2 2 2
Var X x P x
Variance of a Random Variable
Note: so    2 2
x P x E X    2 2 2
E X

10. Mean and Variance of Random Variables
The STANDARD DEVIATION of
a discrete random variable X
is the SQUARE ROOT of its
variance:
    2 2
x P x
The Standard Deviation

11. X P(X)
0 0.05
1 0.15
2 0.35
3 0.45
Find the
variance and
standard
deviation of the
given probability
distribution.
Mean and Variance of Random Variables
Variance of a Random Variable

12. A study conducted by a TV station
showed the number of televisions per
household and the corresponding
probabilities for each. Find the mean,
variance, and standard deviation.
Mean and Variance of Random Variables
Mean of a Random Variable

13. a.An investment in Project A will
result in a loss of P 26,000 with
probability 0.30, break even
with probability 0.50, or result
in a profit of P 68,000 with
probability 0.20. What is the
expected value of this
investment?
Mean and Variance of Random Variables
Mean of a Random Variable

14. b. Meanwhile, an investment in
Project B will result in a loss of
P 71,000 with probability 0.20,
break even with probability
0.65, or result in a profit of
P 143,000 with probability
0.15. Which investment is
better: A or B?
Mean and Variance of Random Variables
Mean of a Random Variable

15. In the Mr. and Ms. CABT Raffle
Draw, there is one winner of P
10,000, one winner of P 5000, and
five winners of P1000. One
thousand tickets are sold at P10
each. Find the expectation if a
person buys one ticket.
Mean and Variance of Random Variables
Mean of a Random Variable

16. In the Mr. and Ms. CABT Raffle Draw, there is one
winner of P 10,000, one winner of P 5000, and five
winners of P1000. One thousand tickets are sold at
P10 each. Find the expectation if a person buys one
ticket.
Mean and Variance of Random Variables
Mean of a Random Variable
Prize 0 1,000 5,000 10,000
Gain (X) 10 990 4,990 9,990
No. of tickets 993 5 1 1
Probability 0.993 0.005 0.001 0.001

17. In the Mr. and Ms. CABT Raffle Draw, there is one
winner of P 10,000, one winner of P 5000, and five
winners of P1000. One thousand tickets are sold at
P10 each. Find the expectation if a person buys one
ticket.
Mean and Variance of Random Variables
Mean of a Random Variable
Gain X 10 990 4,990 9,990
P(X) 0.993 0.005 0.001 0.001
The following is the pdf:

18. JM the financial adviser suggests that his
client Manay Ehljie select one of two types
of bonds in which to invest her P 5,000.
Bond X pays a return of 4% and has a default
rate of 2%. Bond Y has a return of 2.5% and
a default rate of 1%. Find the expected rate
of return and decide which bond would be a
better investment. When the bond defaults,
the investor loses all the investment.
Mean and Variance of Random Variables
Mean of a Random Variable

19. JM the financial adviser suggests that his client Manay Ehljie
select one of two types of bonds in which to invest her P 5,000.
Bond X pays a return of 4% and has a default rate of 2%. Bond Y
has a return of 2.5% and a default rate of 1%. Find the expected
rate of return and decide which bond would be a better
investment. When the bond defaults, the investor loses all the
investment.
Mean and Variance of Random Variables
Mean of a Random Variable
Amount of return for X:
  5,000 0.04 200
Amount of return for Y:
  5,000 0.025 125
The probability of obtaining a return
(that is, NOT defaulting):
For X: 100%  2% = 98% or 0.98
For Y: 100%  2.5% = 97% or 0.975

20. JM the financial adviser suggests that his client Manay Ehljie
select one of two types of bonds in which to invest her P 5,000.
Bond X pays a return of 4% and has a default rate of 2%. Bond Y
has a return of 2.5% and a default rate of 1%. Find the expected
rate of return and decide which bond would be a better
investment. When the bond defaults, the investor loses all the
investment.
Mean and Variance of Random Variables
Mean of a Random Variable
Return X 5,000 200
P(X) 0.02 0.98
Return Y 5,000 125
P(Y) 0.01 0.975

21. Do you have any
QUESTIONs?

23. WhenIlisten,Ihear.
WhenIsee,Iremember.
WhenIdo,Iunderstand.
Final Words….

24. Thank you!