Random Variables and Probabiity Distribution

TOPIC 1: RANDOM VARIABLES
AND PROBABILITY
DISTRIBUTION
PREPARED BY: JESSA R. ALBIT

LESSON 1: EXPLORING RANDOM VARIABLES
LESSON OBJECTIVES:
AT THE END OF THIS LESSON, YOU ARE EXPECTED TO:
• ILLUSTRATE A RANDOM VARIABLE;
• CLASSIFY RANDOM VARIABLES AS DISCRETE OR CONTINUOUS; AND
• FIND THE POSSIBLE VALUES OF A RANDOM VARIABLE.

MOTIVATION
EXPIREMENT OUTCOMES SAMPLE SPACE
TOSS A COIN ONCE
TOSS A COIN TWICE
ROLL A DIE
EXAM RESULT
GAME RESULT

MOTIVATION
TOSS A COIN ONCE HEAD, TAIL
TOSS A COIN TWICE HH, HT, TH, TT
ROLL A DIE 1, 2, 3, 4, 5, 6
EXAM RESULT PASS, FAIL
GAME RESULT WIN, LOSE

MOTIVATION
TOSS A COIN ONCE HEAD, TAIL 𝑆 = 𝐻𝐸𝐴𝐷, 𝑇𝐴𝐼𝐿
TOSS A COIN TWICE HH, HT, TH, TT
𝑆
= HH, HT, TH, TT
ROLL A DIE 1, 2, 3, 4, 5, 6
𝑆
= 1, 2, 3, 4, 5, 6
EXAM RESULT PASS, FAIL 𝑆 = PASS, FAIL
GAME RESULT WIN, LOSE 𝑆 = WIN, LOSE

DEFECTIVE CELLPHONES
Suppose three cell phones are tested at random. We want to find out
the number of defective cell phones that occur. Thus, to each outcome
in the same space we shall assign a value. These are 0, 1, 2, or 3. If there
is no defective cell phone, we assign the number 0; if there is 1
defective cell phone, we assign the number 1; if there are two defective
cell phones, we assign 2; and 3, if there is three defective cell phones.
The number of defective cell phones is a random variable. The possible
values of this random variable are 0, 1, 2, and 3.

POSSIBLE
OUTCOMES
NOT
DEFECTIVE
NOT
DEFECTIVE
NOT
DEFECTIVE
NNN
DEFECTIVE NND
DEFECTIVE
NOT
DEFECTIVE
NDN
DEFECTIVE NDD
DEFECTIVE
NOT
DEFECTIVE
NOT
DEFECTIVE
DNN
DEFECTIVE DND
DEFECTIVE
NOT
DEFEECTIVE
DDN
DEFECTIVE DDD

LET’S COMPLETE THE TABLE BELOW
POSSIBLE OUTCOMES VALUE OF THE RANDOM VARIABLE X

LET’S COMPLETE THE TABLE BELOW
POSSIBLE OUTCOMES VALUE OF THE RANDOM VARIABLE X
NNN 0
NND 1
NDN 1
DNN 1
NDD 2
DND 2
DDN 2
DDD 3
SO, THE POSSIBLE VALUES OF THE RANDOM VARIABLE X ARE 0, 1, 2, AND 3.

RANDOM VARIABLE
• A RANDOM VARIABLE IS A FUNCTION THAT ASSOCIATES A REAL
NUMBER TO EACH ELEMENT IN THE SAMPLE SPACE. IT IS A VARIABLE
WHOSE VALUES ARE DETERMINED BY CHANCE

RANDOM
VARIABLE
DISCRETE
COUNTABLE
OUTCOMES
COUNT DATA
NUMBER OF
STUDENTS
CONTINUOUS
CONTINUOUS
SCALE
MEASURED DATA
HEIGHTS,
WEIGHTS,
TEMPERATURE

ACTIVITY
GIVEN DISCRETE OR CONTINUOUS
THE PRICE OF A HOUSE.
TIME TO DOWNLOAD A WEBPAGE.
ADVERTISING EXPENDITURES OF A
COMPANY.
STUDENT ENROLLMENT IN A CERTAIN
UNIVERSITY.
WATER TEMPERATURE OF NILE RIVER.

ACTIVITY
GIVEN DISCRETE OR CONTINUOUS
THE PRICE OF A HOUSE. DISCRETE
TIME TO DOWNLOAD A WEBPAGE. CONTINUOUS
ADVERTISING EXPENDITURES OF A COMPANY. DISCRETE
STUDENT ENROLLMENT IN A CERTAIN UNIVERSITY. DISCRETE
WATER TEMPERATURE OF NILE RIVER. CONTINUOUS

TRY THIS!
• SUPPOSE THREE COINS ARE TOSSED. LET Y BE THE RANDOM
VARIABLE REPRESENTING THE NUMBER OF TAILS THAT OCCUR.
FIND THE PROBABILITY OF EACH OF THE VALUES OF THE
RANDOM VARIABLE Y.

POSSIBLE
OUTCOMES
Head
H
H HHH
T HHT
T
H HTH
T HTT
Tail
H
H THH
T THT
T
H TTH
T TTT

LESSON 2: CONSTRUCTING PROBABILITY
DISTRIBUTION
LESSON OBJECTIVES:
AT THE END OF THE LESSON, YOU SHOULD BE ABLE TO:
• ILLUSTRATE A PROBABILITY DISTRIBUTION FOR A DISCRETE RANDOM
VARIABLE AND ITS PROPERTIES;
• COMPUTE PROBABILITIES CORRESPONDING TO A GIVEN RANDOM
VARIABLE,; AND
• CONSTRUCT THE PROBABILITY MASS FUNCTION OF A DISCRETE
RANDOM VARIABLE AND ITS CORRESPONDING HISTOGRAM.

DISCRETE PROBABILTY DISTRIBUTION
• A DISCRETE PROBABILITY DISTRIBUTION OR A PROBABILITY MASS
FUNCTION CONSISTS OF THE VALUES A RANDOM VARIABLE CAN
ASSUME AND THE CORRESPONDING PROBABILITIES OF THE VALUES.

USING THE SAME THE DATA FROM LESSON1, LET’S
CONSTRUCT THE PROBABILITY DISTRIBUTION.

SOLUTION:
STEPS SOLUTION
1. DETERMINE THE SAMPLE SPACE. 𝑆 = 𝑁𝑁𝑁, 𝑁𝑁𝐷, 𝑁𝐷𝑁, 𝐷𝑁𝑁, 𝑁𝐷𝐷, 𝐷𝑁𝐷, 𝐷𝐷𝑁, 𝐷𝐷𝐷
2. DO THE TALLY.

STEP 3: ASSIGN PROBABILTY VALUES P(X), TO
EACH OF THE RANDOM VARIABLE.
NUMBER OF DEFECTIVE CELLPHONES
X
P(X)
0 𝑃 0 =
𝑁𝑈𝑀𝐵𝐸𝑅 𝑂𝐹 0
𝑇𝑂𝑇𝐴𝐿 𝑁𝑈𝑀𝐵𝐸𝑅 𝑂𝐹 𝑃𝑂𝑆𝑆𝐼𝐵𝐿𝐸 𝑂𝑈𝑇𝐶𝑂𝑀𝐸
=
1
8
1
3
8
2
3
8
3
1
8
TOTAL
8
8
= 1

HISTOGRAM
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
1 2 3 4
PROBABILITYP(X)
PROBABILY P(X)

PROPERTIES OF A PROBABILITY DISTRIBUTION
1. THE PROBABILTY OF EACH VALUE OF A RANDOM VARIABLE MUST BE BETWEEN
OR EQUAL TO 0 AND 1. IN SYMBOL, WE WRITE IT AS 0 ≤ 𝑃 𝑋 ≤ 1.
2. THE SUM OF THE PROBABILITIES OF ALL VALUES OF THE RANDOM VARIABLE
MUST BE EQUAL TO 1. IN SYMBOL, WE WRITE IT AS 𝑃 𝑋 = 1.

LESSON 3: COMPUTING THE MEAN OF A
DISCRETE PROBABILITY DISTRIBUTION
LESSON OBJECTIVES:
AT THE END OF THIS LESSON, YOU SHOULD BE ABLE TO:
• ILLUSTRATE AND CALCULATE THE MEAN OF A DISCRETE RANDOM
VARIABLE;
• INTERPRET THE MEAN OF A DISCRETE RANDOM VARIABLE; AND
• SOLVE PROBLEMS INVOLVING MEAN OF PROBABILITY
DISTRIBUTIONS.

STEP 1: CONSTRUCT PROBABILITY
DISTRIBUTION
X
P(X)
0
1
8
1
3
8
2
3
8
3
1
8
TOTAL
8
8
= 1

STEP 2:
NUMBER OF DEFECTIVE
CELLPHONES
X
P(X) X* P(X)
0
1
8
0 ×
1
8
= 0
1
3
8
3
8
2
3
8
6
8
3
1
8
3
8
TOTAL
8
8
= 1 𝜇 = 𝑋 ∗ 𝑃 𝑋 =
0
8
+
3
8
+
6
8
+
3
8
=
12
8
= 1.5
STEP 3
SO, THE AVERAGE NUMBER OF DEFECTIVE CELLPHONES THAT
IS TESTED IS 1.5

LESSON 4: COMPUTING THE VARIANCE OF A
DISCRETE PRO
LESSON OBJECTIVES:
AT THE END OF THIS LESSON, YOU SHOULD BE ABLE TO:
• ILLUSTRATE AND CALCULATE THE VARIANCE OF A DISCRETE RANDOM
VARIABLE;
• INTERPRET THE VARIANCEOF A DISCRETE RANDOM VARIABLE; AND
• SOLVE PROBLEMS INVOLVING VARIANCE OF PROBABILITY
DISTRIBUTIONS.

STEP 1: FIND THE MEAN OF THE PROBABILITY
DISTRIBUTION
X
P(X) X* P(X)
0
1
8
0
1
3
8
3
8
2
3
8
6
8
3
1
8
3
8
𝑋 ∗ 𝑃 𝑋 =
0
8
+
3
8
+
6
8
+
3
8
=
12
8
= 1.5

STEP 2-5: FIND THE MEAN OF THE
PROBABILITY DISTRIBUTION
X P(X) X* P(X) 𝑿 − 𝝁 𝑿 − 𝝁 𝟐 𝑿 − 𝝁 𝟐 ∗ 𝑷(𝑿)
0
1
8
0
0-1.5= -
1.5
2.25 0.28
1
3
8
3
8
-1.125 1.27 0.48
2
3
8
6
8
-0.75 0.56 0.21
3
1
8
3
8
-1.125 1.27 0.16
𝜇 = 𝑋 ∗ 𝑃 𝑋 =
0
8
+
3
8
+
6
8
+
3
8
=
12
8
= 1.5
𝜎2
= ( 𝑿 −

RESULT
MEAN: 𝜇 = 𝑋 ∗ 𝑃 𝑋 = 1.5
VARIANCE: 𝜎2
= 𝑿 − 𝝁 𝟐
∗ 𝑷(𝑿) = 1.13
STANDARD DEVIATION: 1.06

Computing Probabilities
Value of X 0 1 2 3
P(x) 1/8 3/8 1/8
Find the following:
1. 𝑃(𝑋 =2)
2. 𝑃(𝑋 <2)
3. 𝑃(𝑋 ≤2)
4. 𝑃(𝑋 > 1)
5. 𝑃(𝑋 = 2 𝑜𝑟 𝑋 = 3)

1. 𝑃(𝑋 =2)
𝑃 𝑥 = 𝑃 𝑥 = 0 + 𝑃 𝑥 = 1 + 𝑃 𝑥 = 2 + 𝑃(𝑥 = 3)
1 =
1
8
+
3
8
+ 𝑃 𝑥 = 2 +
1
8
1 =
5
8
+ 𝑃(𝑥 = 2)
𝑃 𝑥 = 2 = 1 −
5
8
𝑃 𝑥 = 2 =
3
8
The probability of having two defective cellphones is 3/8 or 37.5%.

2. P(x<2)
𝑃 𝑥 < 2 = 𝑃 𝑥 = 1 + 𝑃 𝑥 = 0
𝑃 𝑥 < 2 =
3
8
+
1
8
𝑃 𝑥 < 2 =
4
8
𝑃 𝑥 < 2 = 50%

3. 𝑃 𝑥 ≤ 2
𝑃 𝑥 ≤ 2 = 𝑃 𝑥 = 2 + 𝑃 𝑥 = 1 + 𝑃 𝑥 = 0
𝑃 𝑥 ≤ 2 =
3
8
+
3
8
+
1
8
𝑃 𝑥 ≤ 2 =
7
8
𝑃 𝑥 ≤ 2 = 87.5%

4. 𝑃(𝑋 > 1)
𝑃 𝑋 > 1 = P x = 2 + P x = 3
𝑃(𝑋 > 1 =
3
8
+
1
8
𝑃 𝑋 > 1 =
4
8
𝑃 𝑋 > 1 = 50%

5. 𝑃(𝑋 = 2 𝑜𝑟 𝑋 = 3)
𝑃 𝑋 = 2 𝑜𝑟 𝑋 = 3 = P x = 2 + P x = 3
𝑃(𝑋 = 2 𝑜𝑟 𝑋 = 3) =
3
8
+
1
8
𝑃 𝑋 = 3 𝑜𝑟 𝑋 = 4 =
4
8
𝑃 𝑋 = 3 𝑜𝑟 𝑋 = 4 = 50%

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