CHAPTER I- Part 1.pptx

STATISTICS AND PROBABILITY
Core Subject
Senior High School
JAYSON M. MAGALONG
Subject Teacher

Statistics and probability are sections
of mathematics that deal with data
collection and analysis. Probability is the
study of chance and is a very fundamental
subject that we apply in everyday living,
while statistics is more concerned with how
we handle data using different analysis
techniques and collection methods
 Statistics and Probability Definition

The branches of mathematics concerned
with the laws governing random events,
including the collection, analysis,
interpretation, and display of numerical
data.
 Statistics and Probability Definition

Statistics and Probability
Module 1:
Random Variable and Probability
Distribution
LESSON 1: Random Variable
LESSON 2: Probability Distribution and Its
Properties
LESSON 3: Computing Probability Corresponding
to a Given Random Variable
:

After going through this lesson, you are expected to:
1. illustrates a random variable (discrete and continuous)
(M11/12SP-IIIa-1);
2. distinguished between a discrete and a continuous
random variable
(M11/12SP-IIIa-2);
3. finds the possible values of a random variable (M11/12SP-
IIIa-3);
4. illustrates a probability distribution for a discrete random
variable and its
properties (M11/12SP-IIIa-4); and
5. computes probabilities corresponding to a given random
variable
(M11/12SP-IIIa-6).
 OBJECTIVES

Activity 1: TOSSING COINS!
Directions: Perform the experiment below. After
performing, try to answer the questions that follows. If you
are going to observe on the characteristics of the coin. One
side contains a head, and we will represent that as H, while
the other side which is the tail or T.
Steps:
1. Prepare 3 coins for the activity.
2. Toss the first coin then the second coin and followed
by the last coin.
3. Record the result by writing and indicating whether
it is H or T. If the results of your three tosses for
example is heads, tails, heads, then you will write on
the outcome HTH on the given table below. (Note: If
the outcome is already repeated, do not write anymore
the result. The outcomes

On the given experiment, there are
8 sample spaces and since we are
interested to the number of heads (H) in
each of the possible outcomes, which in
this case are 0, 1, 2, and 3. These are
what we call the RANDOM VARIABLE

LESSON 1: Random Variable
A random variable is a variable whose
value is unknown or a function that assigns
values to each of an experiment's
outcomes.
A random variable is a numerical
description of the outcome of a statistical
experiment
A Random Variable is a capacity that
connects a real number with every
component in the sample space. It is a
variable whose qualities are controlled by
chance.

 TERMINOLOGY
The outcomes are the result of a given
experiment while the sample space is the
set of all possible outcomes of an
experiment.

 A variable X whose value depends on the
outcome of a random process is called a
random variable. A random variable is a
variable whose value is a numerical outcome
of a random phenomenon.
 A random variable is denoted with a capital
letter. The probability distribution of a random
variable X tells what the possible values
of X are and how probabilities are assigned
to those values.
 A random variable can be discrete or
continuous

 Types of Random Variables:
At that point, recognize the two types of
arbitrary factors. These are the discrete and
continuous random variables.
Discrete Random Variables are
variables can take on a finite number of distinct
values. Examples are number of heads
acquired while flipping a coin three times, the
number of kin an individual has, the number of
students present in a study hall at a given time,
and so forth.

Discrete Random Variables are variables that
can take on a finite number of distinct values.
In easier definition, discrete random variable is
a set of possible outcomes that is countable.
Examples are the number of heads acquired
while flipping a coin three times, the number of
defective chairs, the number of boys in the
family, the number of students present in the
online class,

 Types of Random Variables
Continuous Random Variables, then
again, are random variables that take an
interminably uncountable number of
potential values, regularly measurable
amounts. Examples are the height or
weight of an individual, the time an
individual takes for an individual to wash,
time, temperature, item thickness, length,
age, etc.

Continuous Random Variable are
random variables that take an infinitely
uncountable number of potential values,
regularly measurable amounts. Often,
continuous random variables represent
measured data, such as height, weights,
and temperature.

Statement True False
1. There are 4 outcomes if you tossed two coins.
2. If you tossed three coins where X be the random variable
representing the number of tails that occur. The possible
values of the random variable X are 0, 1 and 2
3.The sum of
3
4
+
1
2
=
5
8
4. The sum of 0.25 + 06 +0.36 +0.28=0.95
5.If P(x)=
𝑋+1
6
, the value of P(1) is
1
2
6. If P(x)=
3
𝑥−2
, the value of P(4) is
3
2
Direction: Determine whether the statement is True or False. If the answer is
false, you can modify the statement to make it true.

Example:Number of Defective Computers
Read and analyze the situation given
below:
In a computer laboratory, the teacher wants
to find out if there is a defective computer.
Supposed three computers were tested at random,
she asks one of her Computer System Servicing
students to list all the possible outcomes, such that
D represents the defective computer and N
represents the non-defective computer. Let X be
the random variable for the number of defective
computers. Then, illustrate the probability
distribution of the random variable X.

Based on the above problem, observe,
analyze, and answer the following questions:
a. List the sample space in the given experiment. How
many outcomes are possible?
b. Construct a table showing the number of defective
computers in each outcome and assign this
number to this outcome. What is the value of the
random variable X?
c. Illustrate a probability distribution. What is the
probability value P(X) to each value of the random
variable?
d. What is the sum of the probabilities of all values of
the random variable?
e. What do you notice about the probability of each
value of the random variable?

a. Let D represent the defective computer and N
for the non-defective computer.
The sample space is:
S= {NND, NDN, DNN, DND, DDN, NDD, DDD,
NNN} and there are 8 possible outcomes
b. Count the number of defective computers in
each outcome in the sample space and assign this
number to this outcome. For instance, if you list
NND, the number of defective computers is 1.

Possible Outcomes
Value of the Random Variable X
(number of defective computers)
NND 1
NDN 1
DNN 1
DND 2
DDN 2
NDD 2
DDD 3
NNN 0
There are four possible values of the random
variable X representing the number of defective
computers. The possible values that X can take are
0, 1, 2, and 3.

c. Each of these numbers corresponds to an event
in the sample space S of equally likely outcomes
for this experiment. Since the value of the random
variable X represents the number of defective
computers, X = 0 to (NNN), X = 1 to (NND, NDN,
DNN), X=2 to (DND, DDN, NDD) and X= 3 to
(DDD).
If each of the outcomes is equally likely to occur,
then the probability is:
P (E)=
number of outcomes in the event
number of outcomes in the sample space

Assign probability values P(X) to each value of the
random variable. Since the number of outcomes is 8, the
probability that 0 defective computer will come out is 1/8 or
P (0) = 1/8, the probability that 1 defective computer will
come out is 3/8 or P(1)= is 1/8 , the probability that 2
defective computers will come out is 3/8 or P(2)= 3/8 and
the probability that 3 defective computers will come out is
1/8 or P(3)= 3/8 .
Number of Defective
Computer X
Probability P(X)
0 1
8
or 0.125
1 3
8
or 0.375
2 3
8
or 0.375
3 1
8
or 0.125
X 0 1 2 3
P(X) 1
8
3
8
3
8
1
8

d. by adding all the probabilities
𝟏
𝟖
+
𝟑
𝟖
+
𝟑
𝟖
+
𝟏
𝟖
=
𝟏+𝟑+𝟑+𝟏
𝟖
=
𝟖
𝟖
= 𝟏
you can also use decimals in determining the sum
of the probabilities
0.125 + 0.375 + 0.375 + 0.125 = 1
If you add all the probabilities, the sum
is equal to 1.
e. From the given activity, you can see that the
values of the probability range from 0 to 1.

Properties of discrete probability
distribution
1.The probability of each value of the random
variable must be between or equal to 0 and
1. In symbol 0 < P(X) < 1.
2.The sum of all the probabilities of all values
of the random variable must be equal to 1.
In symbol, we write it as 𝚺P(X) = 1

Ex.1.Determine if the distribution below is a
discrete probability distribution:
X 1 5 7 8 9
PX) 1
3
1
3
1
3
1
3
1
3
The distribution must satisfy that each
probability value P(X) must be
a.) from 0 to 1 and
b) the sum of all the values of the probabilities
must be equal to 1.

By adding all the values of P(X)
Σ P(X)=
1
3
+
1
3
+
1
3
+
1
3
+
1
3
=
𝟓
𝟑
The probability of each value of the random
variable has the same value which is
1
3
and this
value lies between 0 and 1 but the sum of its
probabilities is not equal to 1. Hence, this is not
a probability distribution because ΣP(X) ≠ 1

Ex. 2. Determine if the distribution below is a
discrete probability distribution
You can also express the values of the
probabilities in decimal form.
By adding all the values of P(X)
𝚺 P(X)= 0.35 + 0.25 + 0.28 + 0.12 =1
The probability of each value of the random variable
lies between 0 and 1 and the sum of its probabilities is
equal to 1 or ΣP(X) =1. Therefore, this is a probability
distribution.

Ex. 3. Determine whether the given values can
serve as the values of a probability distribution.
a. P(1) = 0.05, P(2)=1.01, P(3)= 0.2
does not lie between 0 and 1 because P(2)=1.01.
Therefore, this is not a probability distribution.
b. P(1) =
3
20
, P(2)=
7
20
, P(3)=
1
2
3
20
+
7
20
+
1
2
=
3
20
+
7
20
+
10
20
=
20
20
= 1
lies between 0 and 1 and the sum of its probabilities is equal
to 1 or

Ex. 4. Determine whether the following can serve as the
probability distribution of a random variable X.
P(X)=
𝟏
𝟕
for x= 1,2,3,…….9
This means that the value of P(1) to P(9)=
1
7
Illustrating this in a table would give as:
X 1 2 3 4 5 6 7 8 9
P(X) 1
7
1
7
1
7
1
7
1
7
1
7
1
7
1
7
1
7
The probability of each value of the random
variable has the same value which is
1
7
and this value
lies between 0 and 1 but the sum of its probabilities is
9
7
which is not equal to 1. Hence, this is not a
probability distribution because ΣP(X) ≠ 1

b. P(X)=
12
25𝑥
for x= 1,2,3,4
Evaluate P(X), given the value of x
P(1)=
12
25𝑥
=
12
25(1)
=
12
25
= 0.48
P(2)=
12
25𝑥
=
12
25(2)
=
12
50
= 0.24
P(3)=
12
25𝑥
=
12
25(3)
=
12
75
=0.16
P(4)=
12
25𝑥
=
12
25(4)
=
12
100
= 0.12
Illustrating this in a table would give as:
X 1 2 3 4
P(X) 0.48 0.24 0.16 0.12

The probability of each value of the
random variable lies between 0 and 1 and
the sum of its probabilities is equal to 1 or
ΣP(X) =1. Therefore, this is a probability
distribution.

Independent Activity 1
A. Determine whether the distribution represents a
probability distribution or not. Explain your answer.
1 X 1 5 7 8
P(X)
1
4
1
8
1
4
1
8
2 X 0 2 3 4 6
P(X)
1
6
1
6
1
6
1
3
1
6
3 X 1 3 5 7
P(X) 0.35 0.25 0.22 0.12
4. P(1)=0.42, P(2)=0.31, P(3)=0.37
5. P(1)=
𝟏𝟐
𝟑𝟓
, P(2)=
𝟖
𝟑𝟓
, P(3)=
𝟑
𝟕
B. Determine whether the following can serve as the
probability distribution of a random variable X.
1. P(X)=
1
7
for x= 1,2,3,…….7
2. P(X)=
𝑥−2
3
for x= 1,2,3,4,5

Computing
Probability
Corresponding to
a Given Random
Variable

You have learned that the likelihood of
winning in a lotto draw, number of winnings in a
gamble, number of heads that come out in a toss of
a coin, and other game of chance can be estimated
using probability. But do you know that it is not only
for the game of chance? We also use this in
business, economics, engineering, and other real-
life situations. A data needs to determine the
probabilities to make decisions and draw a
conclusion. Most of the time, you won't
perform actual probability problems, but
you'll use probability to make a judgment and
determine the best course of action.

Find the probability of the following events.
Event (E)
Probability P
(E)
a. Getting an odd number in
a single roll of a die
b. Getting an ace when a
card is drawn from a
deck
c. Getting a number greater
than 2 in a single
roll of a die
d. Getting a red queen when
a card is drawn
from a deck
e. Getting doubles when two
dice are rolled
f. Getting a sum of 5 or a
sum of 9 when two
dice are rolled

The probability for each event will be:
The event of getting an odd number has three
outcomes: 1, 3, or 5. Since there are 6 faces in a die, then
the probability of getting an odd number is
3
6
or
1
2
.
The event of getting an ace when a card is drawn
from a deck is 4. Since there are 52 faces in a deck of
cards, then the probability of getting an ace is
4
52
or
1
13
.
The event of getting a number greater than 2 in a
single roll of a die has four outcomes: 3, 4, 5, and 6. Since
there are 6 faces in a die, then the probability of getting a
number greater than 2 is
4
6
or
2
3
.

The event of getting a red queen when a card is
drawn from a deck is 2. Since there are 52 faces in a deck of
cards, then the probability of getting a red queen is
2
52
or
1
26
.
The event of getting doubles when two dice are rolled
has 6 outcomes: {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}.Since
there are 36 outcomes in rolling two dice, then the probability
of getting doubles is
6
36
or
1
6
.
The event of getting a sum of 5 when two dice are
rolled has 4 outcomes:{(1,4), (4,1),(2,3),(3,2)}.Hence, the
probability is
4
36
.The event of getting a sum of 9 when two
dice are rolled has 4 outcomes: {(4,5),(5,4),(3,6),(6,3)}.
Hence, the probability is
4
36
.

The probability of getting a sum of 5 or a
sum of 9 when two dice are rolled is
P(sum of 5 or sum of 9)
=
number of pairs with sums of 5
total number of outcomes
+
number of pairs with sums of 9
total number of outcomes
=
4
36
+
4
36
=
8
36
or
2
9

Golden’s bakery is
known for its famous
Filipino delicacies. Among
these foods which is a
native delicious food
called “kakanin” is a
“leche puto”. The
bakeshop owner recorded
the number of boxes of
“leche puto” that were
delivered each day. The
number of boxes
delivered for 10 days is
shown below.
Day
Number of
Boxes(X)
1 35
2 37
3 50
4 45
5 37
6 45
7 40
8 42
9 45
10 42

1. What is the probability that 40 or more boxes will
be delivered on a particular day?
2. What is the probability that the number of boxes
delivered will be at least 37 but not more than 50?
3. What is the probability that at most 40 boxes will
be delivered on a particular day?
4. Find P (X≤ 45)
5. Find P (40) + P (50)

To answer the given question, you have to construct
first the probability distribution. Let X the value of the
random variable represented by the number of boxes of
“leche puto”. The probability distribution is shown below.
Number of Boxes X Probability P(X)
35 1
10
37 1
5
40 1
10
42 1
5
45 3
10
50 1
10

1. The probability that 40 or more boxes will be
sold in a particular day means P (X≥ 40).
This means that you have to add P(X=40),
P(X=42), P (X= 45), and (X= 50).
P (X≥ 40) = P(40) + P(42) + P(45) + P(50)
then, substitute its corresponding probability
=
1
10
+
1
5
+
3
10
+
1
10
=
1
10
+
2
10
+
3
10
+
1
10
=
7
10
or 0.7

P (37≤ X< 50) = P(37) + P(40) +P(42) + P(45),
=
1
5
+
1
10
+
1
5
+
3
10
=
2
10
+
1
10
+
2
10
+
3
10
=
8
10
=
4
5
or 0.8
2. The probability that the number of boxes
delivered will be at least 37 but not more than 50
means P (37≤ X< 50).
Hence, the values included are P(37), P(40),
P42) and P(45), then substitute its
corresponding probability

3. The probability that at most 40 boxes will
be delivered in a particular day means P (X≤
40) so the values of X are P(40), P(37) and
P(35)
P (X≤ 40) = P(40) + P(37) + P(35)
=
1
10
+
1
5
+
1
10
=
1
10
+
2
10
+
1
10
=
4
10
=
2
5
or 0.40

4. Find P (X≤ 45) = P(45) + P(42) + P(40) + P(37)
+P(35)
=
3
10
+
1
5
+
1
10
+
1
5
+
1
10
=
3
10
+
2
10
+
1
10
+
2
10
+
1
10
=
9
10
or 0.90
Other solution:
P (X≤ 45) = 1-P(50)
= 1-
1
10
=
9
10

5. Find P (40) + P (50) =
1
10
+
1
10
=
2
10
=
1
5
or 0.10

1.P(X≥ 3)
Number of Cell
Phones Sold X
Probability
P(X)
1 0.15
2 0.10
3 0.25
4 0.30
5 0.20

2.P(x≤4)
Number of Cell
Phones Sold X
Probability
P(X)
1 0.15
2 0.10
3 0.25
4 0.30
5 0.20

3. P(1≤X≤4)
Number of Cell
Phones Sold X
Probability
P(X)
1 0.15
2 0.10
3 0.25
4 0.30
5 0.20

4.P(2) + P(4)
Number of Cell
Phones Sold X
Probability
P(X)
1 0.15
2 0.10
3 0.25
4 0.30
5 0.20

Number of Cell
Phones Sold X
Probability
P(X)
1 0.15
2 0.10
3 0.25
4 0.30
5 0.20
5. FIND P(ODD)

Analyze and solve the following problems:
1. Box A and B contain numbers 1,2 3, and 4. The following
is the probability distribution of the sum when one number
from each box is taken at a time with replacement.
X 2 3 4 5 6 7 8
P(X) 1
16
1
8
3
16
1
4
3
16
1
8
1
16
a. Find P(even)
b. Find P (X≥ 5)
c. Find the probability that X assumes a value of less
than 4.
d. Find the probability that X assumes a value
greater than 6.

CHAPTER I- Part 1.pptx

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