Module Code: PASAY-SP11-S2-Q1-W3-D1
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
DEPARTMENT OF EDUCATION- NATIONAL CAPITAL REGION
SCHOOLS DIVISION OF PASAY CITY
MODULE IN STATISTICS AND PROBABILITY
Second Semester / First Quarter / Week 3 / Day 1
Objective: Solves problems involving mean and variance of probability distributions.
YOUR LESSON FOR TODAY:
•
Solving Problems Involving Mean and Variance of Probability Distributions
RECALL:
You have learned about mean and variance of discrete random variable from the previous
lessons. As a review, let us look back to these:
Formula for the Mean of the Probability Distribution
The mean of a random variable with a discrete probability distribution is:
𝜇 = 𝑋1 • 𝑃(𝑋1 ) + 𝑋2 • 𝑃(𝑋2 ) + 𝑋3 • 𝑃(𝑋3 ) + ⋯ , +𝑋𝑛 • 𝑃(𝑋𝑛 ) or
𝜇 = ∑ 𝑋 • 𝑃(𝑋)
where:
𝑋1 , 𝑋2 , 𝑋3 , … , 𝑋𝑛
are the values of the random variable 𝑋; and
𝑃(𝑋1 ), 𝑃(𝑋2 ), 𝑃(𝑋3 ), … , 𝑃(𝑋𝑛 )
are the corresponding probabilities.
Formula for the Variance and Standard Deviation of a Discrete Probability Distribution
The variance of a discrete probability distribution is given by the formula:
𝜎 2 = ∑(𝑋 − 𝜇)2 • 𝑃(𝑋)
The standard deviation of a discrete probability distribution is given by the formula:
𝜎 = √∑(𝑋 − 𝜇)2 • 𝑃(𝑋)
Alternative formula for the variance and standard deviation of a discrete probability
distribution:
𝜎 2 = ∑ 𝑋2 • 𝑃(𝑋) − 𝜇2
𝜎 = √∑ 𝑋2 • 𝑃(𝑋) − 𝜇2
where:
𝑋
= value of the random variable
𝑃(𝑋)
= probability of a random variable 𝑋
𝜇
= mean of the probability distribution
Page 1 of 20
Module Code: PASAY-SP11-S2-Q1-W3-D1
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
Let us try to solve these problems involving mean of probability distributions.
Example 1: Consider rolling a single die. What is the average number of spots that would appear?
Solutions:
Rolling a single die has 6 possible outcomes. Those are 1, 2, 3, 4, 5 and 6. Thus,
1
6
Sample Space : {1, 2, 3, 4, 5, 6}. The probability of getting a 1 is , getting a 2 is
1
6
and also
getting 3, 4, 5 and 6. Let us construct the probability distribution with the corresponding
probabilities of each possible outcome.
𝑋
1
2
3
4
5
6
𝑃(𝑋)
1
6
1
6
1
6
1
6
1
6
1
6
To compute the mean, we will be using the formula. Take note that 𝑋 is the value of the random
variable and 𝑃(𝑋) is its corresponding probability. Thus,
𝜇 = 𝑋1 • 𝑃(𝑋1 ) + 𝑋2 • 𝑃(𝑋2 ) + 𝑋3 • 𝑃(𝑋3 ) + ⋯ , +𝑋𝑛 • 𝑃(𝑋𝑛 )
1
1
1
1
1
1
6
6
6
6
6
6
𝜇 =1•( )+2•( )+3•( )+4•( )+5•( )+6•( )
𝜇=
1 2 3 4 5 6 21
+ + + + + =
= 3.5
6 6 6 6 6 6
6
Thus, the average number of spots that would appear is 3.5.
Example 2: Lian is paid 20 𝑃ℎ𝑝 whenever the results of tossing two coins are both heads but pays
10 𝑃ℎ𝑝 whenever the results are not both heads. What is her expected gain per toss?
Solutions: Let 𝑋 be the random variable defined. There are 4 outcomes in tossing two coins, in which
only 1 is a HH. The other results are HT, TH and TT. The probability of both heads is
3
4
1
4
while the
probability of not both heads is .
To determine her expected gain per toss, we will be using the concept in computing the mean of a
probability distribution. Let’s construct the table.
𝑋
Php20(both heads)
-10Php(not both heads)
𝑃(𝑋)
1
4
3
4
1
4
20 𝑃ℎ𝑝 will be paid if HH appears with a corresponding probability of , and she will pay 10 𝑃ℎ𝑝 if not
both heads that is why it is -10. Using the formula,
1
4
3
4
𝜇 = 20 • ( ) + (−10) • ( ) = 5 − 7.5 = −2.5
This means that Lian would lose 2.50 𝑃ℎ𝑝 per toss.
Page 2 of 20
Module Code: PASAY-SP11-S2-Q1-W3-D1
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
Now let us try to solve these problems involving variance of probability distributions.
Example 1: The number of pizzas sold per day at a pizza shop, with its corresponding probabilities,
is shown in the table. Find the variance and standard deviation of the probability distribution.
𝑋
19
20
21
22
23
𝑃(𝑋)
0.20
0.20
0.30
0.20
0.10
Solution. Variance:
To solve this, we will be using the formula of the variance, 𝜎 2 = ∑ 𝑋2 • 𝑃(𝑋) − 𝜇2
Step 1. Find the mean of the probability distribution.
𝜇 = 19 • (20) + 20 • (0.20) + 21 • (0.30) + 22 • (0.20) + 23 • (0.10) = 20.8
Step 2. Multiply the square of the value of the random variable 𝑋 by its corresponding probability.
𝑋
𝑃(𝑋)
𝑋 • 𝑃(𝑋)
𝑋2 • 𝑃(𝑋)
19
0.20
3.8
192 • 0.20 = 72.2
20
0.20
4
202 • 0.20 = 80
21
0.30
6.3
212 • 0.30 = 132.3
22
0.20
4.4
222 • 0.20 = 96.8
23
0.10
2.3
232 • 0.10 = 52.9
= 20.8
∑ 𝑋2 • 𝑃(𝑋) = 434.2
Step 3. Get the sum of the results obtained in Step 2.
Step 4. Subtract the square of the mean from the results obtained in Step 3 to get the variance.
𝜎 2 = ∑ 𝑋2 • 𝑃(𝑋) − 𝜇2
𝜎 2 = 434.4 − (20.8)2 = 1.56
Solution. Standard deviation
Get the square root of the variance to get the standard deviation.
𝜎 = √1.56 = 1.25
The variance of the probability distribution is 1.56. The standard deviation is 1.25.
Example 2: Find the variance of the number of monthly absences of Mary based on her previous
records of absences as presented in the probability distribution below.
Number of Absences
Percent
0
25%
1
30%
2
30%
3
15%
Page 3 of 20
Module Code: PASAY-SP11-S2-Q1-W3-D1
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
Solutions: To solve this, we will be using the formula of the variance, 𝜎 2 = ∑ 𝑋2 • 𝑃(𝑋) − 𝜇2
𝑋
𝑃(𝑋)
𝑋 • 𝑃(𝑋)
𝑋2 • 𝑃(𝑋)
0
25%
0
02 • 0.25 = 0
1
30%
0.30
12 • 0.30 = 0.30
2
30%
0.60
22 • 0.30 = 1.2
3
15%
0.45
32 • 0.15 = 1.35
= 1.35
∑ 𝑋2 • 𝑃(𝑋) = 2.85
𝜇 = 0 • (0.25) + 1 • (0.30) + 2 • (0.30) + 3 • (0.20) = 1.35
For variance,
For standard deviation,
𝜎 = √1.03 = 1.01
𝜎 2 = ∑ 𝑋 2 • 𝑃(𝑋) − 𝜇2
𝜎 2 = 2.85 − 1.352 = 1.03
PRACTICE EXERCISE 1: Solve the following problems. Show your complete solution on the space
provided below.
1. The random variable 𝑋, representing the number of nuts in a chocolate bar has the following
probability distribution. Compute the mean.
𝑋
0
1
2
3
4
𝑃(𝑋) 1/10 3/10 3/10 2/10 1/10
2. By investing in a particular stock, Jommel can make 4000 𝑃ℎ𝑝 in a month with a probability
of 0.2 or take a loss of 1000 𝑃ℎ𝑝 with a probability of 0.8. What is his expected gain per month?
PRACTICE EXERCISE 2: Solve the following problem. Show your complete solution on the space
provided below.
The number of inquiries per day by the Office of Admissions in a certain university is shown
below. Find the variance and standard deviation.
Number of Inquiries 𝑋
Probability 𝑃(𝑋)
22
0.08
23
0.19
24
0.36
25
0.25
26
0.07
27
0.05
Page 4 of 20
𝑋 • 𝑃(𝑋)
𝑋2 • 𝑃(𝑋)
Module Code: PASAY-SP11-S2-Q1-W3-D1
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
Learning Summary:
To solve problems involving mean and variance of probability distribution, we use the
following formulas:
where:
𝜇 = ∑ 𝑋 • 𝑃(𝑋)
for the mean,
𝜎 2 = ∑(𝑋 − 𝜇)2 • 𝑃(𝑋)
for the variance,
𝜎 2 = ∑ 𝑋 2 • 𝑃(𝑋) − 𝜇2
and
𝜎 = √∑(𝑋 − 𝜇)2 • 𝑃(𝑋)
𝜎=
√∑ 𝑋 2
• 𝑃(𝑋) −
𝜇2
for the standard
deviation.
𝑋
= value of the random variable
𝑃(𝑋)
= probability of a random variable 𝑋
𝜇
= mean of the probability
distribution
EVALUATION:
The probabilities of a machine manufacturing 0, 1, 2, 3, 4, or 5 defective parts in one day are
0.75, 0.17, 0.04, 0.025, 0.01 and 0.005, respectively. Find the:
a.) Mean of the probability distribution
b.) Variance of the probability distribution
c.) Standard deviation of the probability distribution
Prepared by: ADA MAE RUSSEL S. ESPAÑOLA
Pasay City East High School
References:
Belecina, R. R. (2016). Statistics and Probability. Rex Bookstore Inc. Manila, Philippines
Banignon, Jr. R. B. (2018). Statistics and Probability. Educational Resources Corporation. Quezon City, Philippines
Page 5 of 20
Module Code: PASAY-SP11-S2-Q1-W3-D2
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
DEPARTMENT OF EDUCATION- NATIONAL CAPITAL REGION
SCHOOLS DIVISION OF PASAY CITY
MODULE IN STATISTICS AND PROBABILITY
Second Semester / First Quarter / Week 3 / Day 2
Objectives: Illustrates a normal random variable and its characteristics.
YOUR LESSON FOR TODAY:
•
Normal Random Variable and Its Characteristics
TRY TO DISCOVER:
If you asked people about their height, you will find out that most of the people in a specific
population are of average height. The number of people shorter or taller than the average height is
almost equal, and very small number of people are extremely tall or extremely short.
If you administer a test among 100 students, there will be few high scores and few low scores.
Most of the scores will be found in between the extreme scores. If you make the distribution consist
of very large number of cases and the three measures of central tendency equal (mean, median and
mode), what you have now is what we call the normal distribution or simply normal curve.
Normal Probability Distribution
A normal distribution is a distribution of a
continuous random variable whose graph is a bellshaped curve called the normal curve.
Many
random
variables
are
either
normally
distributed, or at least approximately normally distributed. Examples are height, weight, examination
scores and IQ.
Properties of the Normal Probability Distribution
The normal probability distribution has the following properties:
1. The distribution curve is bell-shaped.
2. The curve is symmetrical about its center.
Symmetry
50%
50%
3. The mean, the median, and the mode coincide at the center.
Page 6 of 20
Module Code: PASAY-SP11-S2-Q1-W3-D2
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
𝑚𝑒𝑎𝑛 = 𝑚𝑒𝑑𝑖𝑎𝑛 = 𝑚𝑜𝑑𝑒
The mean gives the location of the
line of symmetry.
4. The width of the curve is determined by the standard deviation 𝜎 of the distribution.
Standard deviation 𝜎 = 1.9
Standard deviation 𝜎 = 1
5. The tails of the curve flatten out indefinitely along the horizontal axis, always approaching the
axis but never touching it. That is, the curve is asymptotic to the base line.
6. The area under the curve is 1. Thus, it represents the probability or proportion or the percentage
associated with specific sets of measurement values.
Example 1:
a. Which curve has greater mean?
b. Which curve has greater
standard deviation?
Solutions:
a. The line of symmetry of curve 𝐴 occurs at 𝑋 = 5. The line of symmetry of curve 𝐵 occurs at
𝑋 = 9. Curve 𝐵 has a greater mean.
b. Curve 𝐵 is more spread out than curve 𝐴, so curve 𝐵 has the greater standard deviation.
Example 2:
The heights of a fully-grown magnolia
bushes are normally distributed. The curve
represents the distribution. What is the
mean height of a fully grown magnolia
bush?
Solutions:
Observing the given distribution, the line if symmetry is at 𝑋 = 8. Thus, the heights of the magnolia
bushes are normally distributed with a mean height of about 8 feet.
Page 7 of 20
Module Code: PASAY-SP11-S2-Q1-W3-D2
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
Example 3:
A trunk diameter of a certain variety of pine tree is normally distributed with a mean of 𝜇 = 150 𝑐𝑚
and a standard deviation of 𝜎 = 30 𝑐𝑚. Sketch a normal curve that described this distribution.
Solutions:
Step 1. Sketch a normal curve.
Step 2. The mean of 150 goes in the middle.
Step 3. Each standard deviation is a distance of 30 cm.
60
120
90
150
180
210
240
150 − 30 = 120
150 + 30 = 180
120 − 30 = 90
180 + 30 = 210
90 − 30 = 60
210 + 30 = 240
Example 4:
The height of the same variety of pine tree are also normally distributed. The mean height is 𝜇 = 33 𝑚
and the standard deviation is 𝜎 = 3 𝑚. Which normal distribution below summarizes the data?
A
C
B
Solutions:
The mean is 33, it should be found in the center. The standard deviations is 3, so the
distances between the 𝑥 values should be 3. The normal distribution that summarizes the data is B.
Page 8 of 20
Module Code: PASAY-SP11-S2-Q1-W3-D2
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
PRACTICE EXERCISE 1:
Modified True or False. Determine whether each statement is true or false. Write True if the
statement is true, otherwise change the underlined word to make the statement correct.
_______________1.
_______________2.
_______________3.
_______________4.
_______________5.
The normal curve is a bell-shaped distribution.
In a normal curve, the mean, median and mode have equal values.
A normal curve is asymptotic to the y-axis.
The normal curve is symmetric to the x-axis
The total area under the normal curve is 1.
PRACTICE EXERCISE 2:
On the spaces provided before each item, write the correct answers for each of the following
questions.
______________1. Which curve has a greater mean?
______________2. Which curve has a greater standard deviation?
______________3. The length of an adult rabbit is normally distributed. The curve below
represents the distribution. What is the mean length of an adult rabbit?
18
22
26
30
34
38
42
______________4. Based on the data on number 2, what is the standard deviation?
______________5. The weight of adults in Barangay X approaches a normal distribution, and the
curve below represents it. What is the mean weight of an adult in Barangay X?
31.3
38.5
45.7
52.9
Page 9 of 20
60.1
67.3
74.5
Module Code: PASAY-SP11-S2-Q1-W3-D2
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
Learning Summary:
A normal distribution has the following properties:
•
Its curve is bell-shaped.
•
Its curve is symmetric about the y-axis.
•
The mean, median and the mode coincide at the center of the distribution.
•
Its curve is asymptotic with respect to the x-axis.
•
The total area under its curve is 1.
EVALUATION:
Illustrate the normal distribution for each of the following data given.
1.
The mean is 135. The standard
deviation is 12.
2.
The mean is 256. The standard
deviation is 15.
3.
𝜇 = 10 𝜎 = 0.8
4.
𝜇 = 78.9 𝜎 = 2.9
5.
𝜇 = 188.2 𝜎 = 21.1
Prepared by: ADA MAE RUSSEL S. ESPAÑOLA
Pasay City East High School
References:
Belecina, R. R. (2016). Statistics and Probability. Rex Bookstore Inc. Manila, Philippines
Banignon, Jr. R. B. (2018). Statistics and Probability. Educational Resources Corporation. Quezon City, Philippines
Page 10 of 20
Module Code: PASAY-SP11-S2-Q1-W3-D3
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
DEPARTMENT OF EDUCATION- NATIONAL CAPITAL REGION
SCHOOLS DIVISION OF PASAY CITY
MODULE IN STATISTICS AND PROBABILITY
Second Semester / First Quarter / Week 3 / Day 3
Objectives: Identifies regions under the normal curve corresponding to different standard normal
values.
YOUR LESSON FOR TODAY:
• Identifying Regions Under Normal Curve Corresponding to Different Standard Normal Values
TRY TO DISCOVER:
A normally distributed random
variable with a mean µ = 0 and standard
deviation ơ = 1 is called a standard
normal variable. It is presented using
standard normal distribution where the
center of the curve is zero, which is mean
and added one unit from the center to the
right and subtract one unit from the center
to the left. This is shown on the right.
Since the standard normal distribution is normally distributed, 50% of the region below the
curve is found below the mean, while the other 50% is found above the mean.
We can easily find the area of the region
below the standard normal curve using the z-Table
(see z-Table on the last page of this module). This
table gives the area of the region for any value of 𝑧
from −3.99 to 3.99. The value from this table will give
the area of the specific region of the curve to the left
of the given 𝑧-value.
Example 1: Find the area to the left of −2.49.
Solutions:
First, split the given 𝑧-value. The
whole number and the tenths digit (−𝟐. 𝟒) are at the left side of the table while the hundredths
(𝟎. 𝟎𝟗) is at the uppermost of the table. The intersection of these numbers is the area of the normal
curve to the left of the 𝑧-value.
Illustration: 𝒛 = −𝟐. 𝟒𝟗
Answer: The area to the left of −2.49 is 0.0064.
Steps in Finding the Area of the Region Under the Normal Curve Corresponding to Z-Value:
1. Sketch the normal curve and locate the given 𝑧-value.
2. Shade the region given: whether it is below, above of between 𝑧-values.
3. Using the 𝑧-table, find the area of the given 𝑧-value/s.
Page 11 of 20
Module Code: PASAY-SP11-S2-Q1-W3-D3
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
4. Identify the appropriate operation to be used based on step 2:
a. If the region is below or any related terms (to the left, less than), the area is the value
obtained from Step 3.
b. If the region is above or any related terms (to the right, greater than), the area is 1
minus the value obtained in Step 3.
c. If the region is between 𝑧-values, subtract the bigger value by the smaller value
obtained in Step 3.
Example 2: Find the area of the region below 𝑧 = 1.23.
Solutions:
Step 1. Sketch the
normal curve and
locate the given 𝑧value.
Step 2. Shade the
region given:
whether it is
below, above of
between 𝑧-values.
Step 3. Using the
𝑧-table, find the
area of the given 𝑧value.
𝑧 = 1.23
The intersection between 1.2 and 0.03 is 0.8907.
Step 4. Identify
the appropriate
operation to be
used based on
step 2.
The area to be found is the region below, and the value obtained in Step 3
is 0.8907, therefore the area of the region below 𝑧 = 1.23 is 0.8907.
Example 3: Find the area to the right of 𝑧 = 1.23.
Step 1.
Step 2.
Step 3.
The intersection between 1.2 and 0.03 is 0.8907.
Page 12 of 20
Module Code: PASAY-SP11-S2-Q1-W3-D3
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
The area to be found is the region to the right of z, the area will be 1
minus the value obtained in Step 3.
Step 4.
𝐴𝑟𝑒𝑎 = 1 − 0.8907 = 0.1093
Therefore, the area to the right of 𝑧 = 1.23 is 0.1093.
Example 4: Find the area between 𝑧 = 1.23 and 𝑧 = −0.75.
Step 1.
Step 2.
Step 3.
The value that corresponds to 𝑧 = −0.75 is 0.2266 and 𝑧 = 1.23 is 0.8907.
The area to be found is the region between two z-values, the area will be
0.8907 − 0.2266.
Step 4.
𝐴𝑟𝑒𝑎 = 0.8907 − 0.2266 = 0.6641
Therefore, between 𝑧 = 1.23 and 𝑧 = −0.75 is 0.6641.
PRACTICE EXERCISE 1:
Identify the regions under the normal curve of the following standard normal values using
the 𝑧 −table.
1.
2.
3.
4.
5.
𝑧=2
𝑧 = −2.23
𝑧 = 0.18
𝑧 = −0.11
𝑧 = 1.96
_____________________
_____________________
_____________________
_____________________
_____________________
PRACTICE EXERCISE 2:
Find the area of the region indicated in each item. Sketch and label each curve.
Sketch of the Normal Curve
1. to the right of
𝑧 = 1.49
2. below 𝑧 = −1.26
3. above 𝑧 = 0.03
4. between 𝑧 =
−2.11 and 𝑧 =
1.13
5. between 𝑧 =
0.19 and 𝑧 =
2.75
Page 13 of 20
Module Code: PASAY-SP11-S2-Q1-W3-D3
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
Learning Summary:
Steps in Finding the Area of the Region Under the Normal Curve Corresponding to Z-Value:
1.
2.
3.
4.
Sketch the normal curve and locate the given 𝑧-value.
Shade the region given: whether it is below, above of between 𝑧-values.
Using the 𝑧-table, find the area of the given 𝑧-value/s.
Identify the appropriate operation to be used based on step 2:
a. If the region is below or any related terms (to the left, less than), the area is the value
obtained from Step 3.
b. If the region is above or any related terms (to the right, greater than), the area is 1
minus the value obtained in Step 3.
c. If the region is between 𝑧-values, subtract the bigger value by the smaller value
obtained in Step 3.
EVALUATION:
Determine the area of the region indicated in each item. Draw and label each curve.
Sketch of the Normal Curve
1. to the right of 𝑧 =
−2.5
2. below 𝑧 = −0.15
3. above 𝑧 = 1.67
4. between 𝑧 = 0.18 and
𝑧 = 2.3
5. between 𝑧 = −1.88
and 𝑧 = −0.89
Prepared by: ADA MAE RUSSEL S. ESPAÑOLA
Pasay City East High School
References:
Belecina, R. R. (2016). Statistics and Probability. Rex Bookstore Inc. Manila, Philippines
Banignon, Jr. R. B. (2018). Statistics and Probability. Educational Resources Corporation. Quezon City, Philippines
Page 14 of 20
𝒁 − 𝑻𝒂𝒃𝒍𝒆
Page 15 of 20
Module Code: PASAY-SP11-S2-Q1-W3-D4
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
DEPARTMENT OF EDUCATION- NATIONAL CAPITAL REGION
SCHOOLS DIVISION OF PASAY CITY
MODULE IN STATISTICS AND PROBABILITY
Second Semester / First Quarter / Week 3 / Day 4
Objectives: Converts a normal random variable to a standard normal variable and vice versa.
YOUR LESSON FOR TODAY:
•
Converting a Normal Random Variable to a Standard Normal Variable and Vice Versa
TRY TO DISCOVER:
Suppose you scored 40 in a test in Statistics and Probability and the scores are normally
distributed with a mean of 34 and standard deviation of 3, then your score is exactly 2 standard
deviations above the mean. This 2 is called a 𝒛 −score. All values above the mean are positive
𝑧 −scores while all values below the mean are negative 𝑧 −scores . If your score is 31, then your
score if 1 standard deviation below the mean and your 𝑧 −score is −1.
The process of converting a random variable 𝑋 to a standard normal variable or 𝑧 −score is
called “standardizing” or “standardization” of a random variable. It is useful to standardize the
values of a normal distribution by changing them to 𝑧 −scores because:
a. it allows researchers to calculate the probability of a score occurring within a normal
distribution; and
b. enables us to compare two scores that are from different samples.
𝒁 −score or standard score tells us how many standard deviations a value is, away from
the mean. A negative 𝑧 −scores tells that the value is below the mean, while a positive
𝑧 −score tells us that the value is above the mean.
The formula used in converting a random variable 𝑋 to a standard normal variable 𝑧 is
𝑥−𝜇
𝑧=
𝜎
where:
𝑧−
standard normal score, 𝑧 −score
𝑥−
any data value in a normal distribution
𝜇−
mean
𝜎−
standard deviation
Example 1: Given the mean, 𝜇 = 50 and the standard deviation 𝜎 = 4, find the 𝑧 −value that
corresponds to a score 𝑥 = 58.
Solutions:
The 𝑧 −score can be computed using the formula 𝑧 =
Given: 𝜇 = 50
𝜎=4
𝑥 = 58
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𝑥−𝜇
𝜎
.
Module Code: PASAY-SP11-S2-Q1-W3-D4
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
Substitute the variables by its corresponding values then simplify.
𝑥−𝜇
𝑧=
𝜎
58 − 50 8
𝑧=
= =2
4
4
𝑧=2
Thus, the 𝑧 −value that corresponds to the raw score 58 is 2.
Example 2: Suppose IQ scores are normally distributed with a mean of 100 and standard deviation
of 10. If your IQ is 85, what is you 𝑧 −score? (Round off your answer to the nearest hundredths)
Solutions:
The 𝑧 −score can be computed using the formula 𝑧 =
Given: 𝜇 = 100
𝜎 = 10
𝑥−𝜇
.
𝜎
𝑥 = 85
Substitute the variables by its corresponding values then simplify.
𝑥−𝜇
𝑧=
𝜎
85 − 100 15
𝑧=
=
= −1.50
10
10
𝑧 = −1.50
Example 3: The weight of chocolate bars from a particular chocolate factory has a mean of 8 ounces
with standard deviation of 0.1 ounce. What is the z-score corresponding to a weight of 8.17 ounces?
(Round off your answer to the nearest hundredths)
Solutions:
The 𝑧 −score can be computed using the formula 𝑧 =
Given: 𝜇 = 8
𝜎 = 0.1
𝑥−𝜇
𝜎
.
𝑥 = 8.17
Substitute the variables by its corresponding values then simplify.
𝑥−𝜇
𝑧=
𝜎
8.17 − 8 0.17
𝑧=
=
= 1.70
0.1
0.1
𝑧 = −1.70
Converting 𝒛 −score to a normal variable 𝑿:
By manipulating the formula 𝑧 =
𝑥−𝜇
𝜎
, we derive the formula in finding the normal variable 𝑋
given the mean, standard deviation and 𝑧 −score. Thus, the formula in finding the normal variable 𝑋
given the mean, standard deviation and 𝑧 −score is:
𝑥 = 𝑧𝜎 + 𝜇
where:
𝑥−
any data value in a normal distribution
𝑧−
standard normal score, 𝑧 −score
𝜇−
mean
𝜎−
standard deviation
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Module Code: PASAY-SP11-S2-Q1-W3-D4
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
Example 4: If Sharon’s 𝑧 −score on post-assessment is 2, the mean is 10 and the standard
deviation is 2, what was her raw score?
Solutions:
In this example, z-score is given while the normal random variable is unknown. To solve for
the normal random variable 𝑥, multiply the 𝑧 −score 𝑧 by the standard deviation 𝜎, then add
the mean 𝜇, or simply use the formula 𝑥 = 𝑧𝜎 + 𝜇.
Given: 𝑧 = 2
𝜇 = 10
𝜎=2
Substitute the variables by its corresponding values then simplify.
𝑥 = 𝑧𝜎 + 𝜇
𝑥 = (2)(2) + (10) = 4 + 10 = 14
𝑥 = 14
Thus, Sharon’s raw score is 14.
Example 5: The monthly electric bills in a city are normally distributed with a mean of 6,000 𝑃ℎ𝑝
and a standard deviation of 800 𝑃ℎ𝑝. Find the 𝑥 −value corresponding to a 𝑧 −score of 1.60.
Solutions:
In this example, z-score is given while the normal random variable is unknown. To solve for
the normal random variable 𝑥, multiply the 𝑧 −score 𝑧 by the standard deviation 𝜎, then add
the mean 𝜇, or simply use the formula 𝑥 = 𝑧𝜎 + 𝜇.
Given: 𝑧 = 1.60
𝜇 = 6000
𝜎 = 800
Substitute the variables by its corresponding values then simplify.
𝑥 = 𝑧𝜎 + 𝜇
𝑥 = (1.60)(800) + (6000) = 1280 + 6000 = 7280
𝑥 = 7280
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Module Code: PASAY-SP11-S2-Q1-W3-D4
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
PRACTICE EXERCISE 1:
Given the raw score 𝑥, mean 𝜇 and standard deviation 𝜎, determine the 𝑧 −score for each of
the following items. (Round off your answers to the nearest hundredths)
𝒙
𝝁
𝝈
1.
25
40
6
2.
39
30
3
3.
125
105
8
4.
39
45
6
5.
20
26
4
𝒛
PRACTICE EXERCISE 2:
Determine the raw score 𝑋 for each of the following items.
𝒛
𝝁
𝝈
1.
−2.25
218
36
2.
1.11
145
20
3.
3.25
150
15
4.
−1.35
60
5
5.
2.25
70
9
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𝒙
Module Code: PASAY-SP11-S2-Q1-W3-D4
Name: __________________________________________ Track/Strand: _______________
Teacher: _________________________________________ Grade Level: _______________
Learning Summary:
A 𝑧 −score is a measure of the number of standard deviations (𝜎) a particular data value is away
from the mean (𝜇).
The formula used in finding the 𝑧 −score is,
𝑧=
𝑥−𝜇
𝜎
While the formula in finding the raw score 𝑋 is given by,
𝑥 = 𝑧𝜎 + 𝜇
where:
𝑧−
standard normal score, 𝑧 −score
𝑥−
any data value in a normal distribution
𝜇−
mean
𝜎−
standard deviation
Evaluation:
Solve for what is asked in each of the following situations.
1. The scores of students in a midyear examination in Mathematics is normally distributed
with a mean 𝜇 = 34 and a standard deviation 𝜎 = 3. Find the 𝑧 −scores corresponding to
each of the following raw scores:
a. 30
b. 41
c. 31
2. Mayumi and Marikit who are best friends took a college entrance examination. The mean
score and standard deviation of the examination are 115 and 12, respectively. If Mayumi
and Marikit obtained z-scores of 1.5 and -0.95 respectively, what was their scores in the
examination?
a. Mayumi’s score = ____________
b. Marikit’s score = _____________
Prepared by: ADA MAE RUSSEL S. ESPAÑOLA
Pasay City East High School
References:
Belecina, R. R. (2016). Statistics and Probability. Rex Bookstore Inc. Manila, Philippines
Banignon, Jr. R. B. (2018). Statistics and Probability. Educational Resources Corporation. Quezon City, Philippines
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