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Measures of Variability
Grouped Data
Range of Grouped Data
• The range is the simplest measures of variability. The range of a
frequency distribution is simple the difference between the upper
class boundary of the top interval and the lower class boundary of the
bottom interval.
Range =
Upper Class Boundary
of the Highest Interval
-
Lower Class Boundary
of the Lowest Interval
Illustrative Example: solve for the range
Scores in the Second Periodical
Test of 7 – Faith in Mathematics 7
Scores Frequency
46 – 50 1
41 – 45 10
36 – 40 10
31 – 35 16
26 – 30 9
21 - 25 4
Solutions:
Upper Class Limit of the highest Interval = 50
Upper Class Boundary of the Highest Interval = 50 + 0.5 = 50.5
Lower Class Limit of the lowest Interval = 21
Lower Class Boundary of the Lowest Interval = 21 - 0.5 = 20.5
Range =
Upper Class Boundary
of the Highest Interval
-
Lower Class Boundary
of the Lowest Interval
Range = 50.5 – 20.5
Range = 30
Therefore, the range of the given data set is 30.
Variance of Grouped Data ( 𝜎2
)
• Variance is the mean of the square of the deviations from the mean
of a frequency distribution.
• For large quantities, the variance is computed using frequency and
midpoint value for each interval, the deviation and its square, and the
product of the frequency and the squared deviation.
• To find the variance of the grouped data, use the formula:
𝜎2
=
𝑓(𝑋 − 𝑥)2
𝑓 − 1
Where:
f = class frequency
X = Class Mark
𝑥 = class mean
∑f = total number of frequency
In calculating the variance, do the following steps:
1. Prepare a frequency distribution with appropriate class intervals and
write the corresponding frequency (f).
2. Get the midpoint (X) of each class interval.
3. Multiply frequency (f) and the midpoint (X) of each interval to get fX.
4. Add fX of each interval to get ∑fX.
5. Compute the mean using 𝑥 =
𝑓𝑋
𝑓
.
6. Calculate the deviation (X - 𝑥) by subtracting the mean from each
midpoint.
7. Square the deviation of each interval to get (𝑋 − 𝑥)2.
8. Multiply frequency (f) and (𝑋 − 𝑥)2. Fond the sum of each product to
get 𝑓(𝑋 − 𝑥)2 .
9. Calculate the variance using the formula 𝜎2
=
𝑓(𝑋 − 𝑥)2
𝑓 − 1
Find the variance of
the given data set:
Scores in the Second Periodical
Test of 7 – Faith in Mathematics 7
Scores Frequency
46 – 50 1
41 – 45 10
36 – 40 10
31 – 35 16
26 – 30 9
21 - 25 4
In calculating the variance, do the following steps:
1. Prepare a frequency distribution with appropriate
class intervals and write the corresponding frequency
(f).
2. Get the midpoint (X) of each class interval in column 2.
3. Multiply frequency (f) and the midpoint (X) of each
interval to get fX.
4. Add fX of each interval to get ∑fX.
5. Compute the mean using 𝑥 =
𝑓𝑋
𝑓
.
6. Calculate the deviation (X - 𝑥) by subtracting the mean
from each midpoint.
7. Square the deviation of each interval to get (𝑋 − 𝑥)2.
8. Multiply frequency (f) and (𝑋 − 𝑥)2. Fond the sum of
each product to get 𝑓(𝑋 − 𝑥)2 .
9. Calculate the variance using the formula
𝜎2 =
𝑓(𝑋 − 𝑥)2
𝑓 − 1
Solution
Scores Frequency
Class Mark
(X)
fX (X - 𝑥) (X − 𝑥) 2
f(X − 𝑥) 2
46 – 50 1 48 48 13.4 179.56 179.56
41 – 45 10 43 430 8.4 70.56 705.6
36 – 40 10 38 380 3.4 11.56 115.6
31 – 35 16 33 528 -1.6 2.56 40.96
26 – 30 9 28 252 - 6.6 43.56 392.04
21 - 25 4 23 92 - 11.6 134.56 538.24
i = 5 𝒇 = 50
𝒇𝑿 = 𝟏, 𝟕𝟑𝟎 𝒇 𝑿 − 𝒙 𝟐 = 𝟏, 𝟗𝟕𝟐
Mean ( 𝑥) =
𝑓𝑋
𝑓
=
1,730
50
= 34.60
𝜎2 =
𝑓(𝑋 − 𝑥)2
𝑓 − 1
𝜎2
=
1.972
50 − 1
𝜎2
=
1.972
49
𝜎2 = 40.2448 ≈ 40.24
Therefore, the variance (𝝈 𝟐)
is 40.24
Standard Deviation
• The standard deviation is considered the best indicator of the degree
of dispersion among the measures of variability because it represents
an average variability of the distribution.
• Given the set of data, the smaller the range, the smaller the standard
deviation, the less spread is the distribution.
• To get the value of the distribution (s), get the square root of the
variance (𝜎2).
𝑠 = 𝜎2
Solution
Scores Frequency
Class Mark
(X)
fX (X - 𝑥) (X − 𝑥) 2
f(X − 𝑥) 2
46 – 50 1 48 48 13.4 179.56 179.56
41 – 45 10 43 430 8.4 70.56 705.6
36 – 40 10 38 380 3.4 11.56 115.6
31 – 35 16 33 528 -1.6 2.56 40.96
26 – 30 9 28 252 - 6.6 43.56 392.04
21 - 25 4 23 92 - 11.6 134.56 538.24
i = 5 𝒇 = 50
𝒇𝑿 = 𝟏, 𝟕𝟑𝟎 𝒇 𝑿 − 𝒙 𝟐 = 𝟏, 𝟗𝟕𝟐
Mean ( 𝑥) =
𝑓𝑋
𝑓
=
1,730
50
= 34.60
𝜎2 =
𝑓(𝑋 − 𝑥)2
𝑓 − 1
𝜎2
=
1.972
50 − 1
𝜎2
=
1.972
49
𝝈 𝟐 = 𝟒𝟎. 𝟐𝟒𝟒𝟖 ≈ 𝟒𝟎. 𝟐𝟒
𝑠 = 𝜎2
𝑠 = 40.24
𝒔 = 𝟔. 𝟑𝟒
Complete the frequency distribution table by finding the unknown values.
Scores of Grade 7 Avocado Students in 4th Periodical Test in Mathematics
Score Frequency
(f)
Class
Mark
(X)
fX (X - 𝑥) (X − 𝑥) 2
f(X − 𝑥) 2
46 – 50 2
41 – 45 9
36 – 40 13
31 – 35 11
26 – 30 10
21 – 25 5
i =
𝑓 = 𝑓𝑋 = 𝑓 𝑋 − 𝑥 2
=
Questions:
1. How did you determine the
unknown values in the frequency
distribution table?
2. What is the class size?
3. What is the 𝑓𝑋 ?
4. What is the value of the mean in the
given distribution table?
5. What is the upper class boundary of
the top interval? What about the
lower class boundary of the bottom
interval?
6. What is the range
7. What is the variance of the given
distribution table?
8. How would you find the variance?
9. What is the standard deviation?
10.How would you solve for the
standard deviation?
Calculate the range, variance, and standard deviation of the Weekly
Allowance of students in Binago School of Fisheries. Write your
complete solutions and answers on a sheet of paper.
Weekly allowance of Students in Binago School of Fisheries
Weekly Allowance
(in Pesos)
Frequency Class Mark (X) fX (X - 𝑥) (X − 𝑥) 2
f(X − 𝑥) 2
500 – 549 2
450 – 499 3
400 – 449 1
350 – 399 3
300 – 349 4
250 – 299 14
200 – 249 12
150 – 199 21
100 - 149 10

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Measures of variability grouped data

  • 2. Range of Grouped Data • The range is the simplest measures of variability. The range of a frequency distribution is simple the difference between the upper class boundary of the top interval and the lower class boundary of the bottom interval. Range = Upper Class Boundary of the Highest Interval - Lower Class Boundary of the Lowest Interval
  • 3. Illustrative Example: solve for the range Scores in the Second Periodical Test of 7 – Faith in Mathematics 7 Scores Frequency 46 – 50 1 41 – 45 10 36 – 40 10 31 – 35 16 26 – 30 9 21 - 25 4 Solutions: Upper Class Limit of the highest Interval = 50 Upper Class Boundary of the Highest Interval = 50 + 0.5 = 50.5 Lower Class Limit of the lowest Interval = 21 Lower Class Boundary of the Lowest Interval = 21 - 0.5 = 20.5 Range = Upper Class Boundary of the Highest Interval - Lower Class Boundary of the Lowest Interval Range = 50.5 – 20.5 Range = 30 Therefore, the range of the given data set is 30.
  • 4. Variance of Grouped Data ( 𝜎2 ) • Variance is the mean of the square of the deviations from the mean of a frequency distribution. • For large quantities, the variance is computed using frequency and midpoint value for each interval, the deviation and its square, and the product of the frequency and the squared deviation. • To find the variance of the grouped data, use the formula: 𝜎2 = 𝑓(𝑋 − 𝑥)2 𝑓 − 1 Where: f = class frequency X = Class Mark 𝑥 = class mean ∑f = total number of frequency
  • 5. In calculating the variance, do the following steps: 1. Prepare a frequency distribution with appropriate class intervals and write the corresponding frequency (f). 2. Get the midpoint (X) of each class interval. 3. Multiply frequency (f) and the midpoint (X) of each interval to get fX. 4. Add fX of each interval to get ∑fX. 5. Compute the mean using 𝑥 = 𝑓𝑋 𝑓 . 6. Calculate the deviation (X - 𝑥) by subtracting the mean from each midpoint. 7. Square the deviation of each interval to get (𝑋 − 𝑥)2. 8. Multiply frequency (f) and (𝑋 − 𝑥)2. Fond the sum of each product to get 𝑓(𝑋 − 𝑥)2 . 9. Calculate the variance using the formula 𝜎2 = 𝑓(𝑋 − 𝑥)2 𝑓 − 1
  • 6. Find the variance of the given data set: Scores in the Second Periodical Test of 7 – Faith in Mathematics 7 Scores Frequency 46 – 50 1 41 – 45 10 36 – 40 10 31 – 35 16 26 – 30 9 21 - 25 4 In calculating the variance, do the following steps: 1. Prepare a frequency distribution with appropriate class intervals and write the corresponding frequency (f). 2. Get the midpoint (X) of each class interval in column 2. 3. Multiply frequency (f) and the midpoint (X) of each interval to get fX. 4. Add fX of each interval to get ∑fX. 5. Compute the mean using 𝑥 = 𝑓𝑋 𝑓 . 6. Calculate the deviation (X - 𝑥) by subtracting the mean from each midpoint. 7. Square the deviation of each interval to get (𝑋 − 𝑥)2. 8. Multiply frequency (f) and (𝑋 − 𝑥)2. Fond the sum of each product to get 𝑓(𝑋 − 𝑥)2 . 9. Calculate the variance using the formula 𝜎2 = 𝑓(𝑋 − 𝑥)2 𝑓 − 1
  • 7. Solution Scores Frequency Class Mark (X) fX (X - 𝑥) (X − 𝑥) 2 f(X − 𝑥) 2 46 – 50 1 48 48 13.4 179.56 179.56 41 – 45 10 43 430 8.4 70.56 705.6 36 – 40 10 38 380 3.4 11.56 115.6 31 – 35 16 33 528 -1.6 2.56 40.96 26 – 30 9 28 252 - 6.6 43.56 392.04 21 - 25 4 23 92 - 11.6 134.56 538.24 i = 5 𝒇 = 50 𝒇𝑿 = 𝟏, 𝟕𝟑𝟎 𝒇 𝑿 − 𝒙 𝟐 = 𝟏, 𝟗𝟕𝟐 Mean ( 𝑥) = 𝑓𝑋 𝑓 = 1,730 50 = 34.60 𝜎2 = 𝑓(𝑋 − 𝑥)2 𝑓 − 1 𝜎2 = 1.972 50 − 1 𝜎2 = 1.972 49 𝜎2 = 40.2448 ≈ 40.24 Therefore, the variance (𝝈 𝟐) is 40.24
  • 8. Standard Deviation • The standard deviation is considered the best indicator of the degree of dispersion among the measures of variability because it represents an average variability of the distribution. • Given the set of data, the smaller the range, the smaller the standard deviation, the less spread is the distribution. • To get the value of the distribution (s), get the square root of the variance (𝜎2). 𝑠 = 𝜎2
  • 9. Solution Scores Frequency Class Mark (X) fX (X - 𝑥) (X − 𝑥) 2 f(X − 𝑥) 2 46 – 50 1 48 48 13.4 179.56 179.56 41 – 45 10 43 430 8.4 70.56 705.6 36 – 40 10 38 380 3.4 11.56 115.6 31 – 35 16 33 528 -1.6 2.56 40.96 26 – 30 9 28 252 - 6.6 43.56 392.04 21 - 25 4 23 92 - 11.6 134.56 538.24 i = 5 𝒇 = 50 𝒇𝑿 = 𝟏, 𝟕𝟑𝟎 𝒇 𝑿 − 𝒙 𝟐 = 𝟏, 𝟗𝟕𝟐 Mean ( 𝑥) = 𝑓𝑋 𝑓 = 1,730 50 = 34.60 𝜎2 = 𝑓(𝑋 − 𝑥)2 𝑓 − 1 𝜎2 = 1.972 50 − 1 𝜎2 = 1.972 49 𝝈 𝟐 = 𝟒𝟎. 𝟐𝟒𝟒𝟖 ≈ 𝟒𝟎. 𝟐𝟒 𝑠 = 𝜎2 𝑠 = 40.24 𝒔 = 𝟔. 𝟑𝟒
  • 10. Complete the frequency distribution table by finding the unknown values. Scores of Grade 7 Avocado Students in 4th Periodical Test in Mathematics Score Frequency (f) Class Mark (X) fX (X - 𝑥) (X − 𝑥) 2 f(X − 𝑥) 2 46 – 50 2 41 – 45 9 36 – 40 13 31 – 35 11 26 – 30 10 21 – 25 5 i = 𝑓 = 𝑓𝑋 = 𝑓 𝑋 − 𝑥 2 = Questions: 1. How did you determine the unknown values in the frequency distribution table? 2. What is the class size? 3. What is the 𝑓𝑋 ? 4. What is the value of the mean in the given distribution table? 5. What is the upper class boundary of the top interval? What about the lower class boundary of the bottom interval? 6. What is the range 7. What is the variance of the given distribution table? 8. How would you find the variance? 9. What is the standard deviation? 10.How would you solve for the standard deviation?
  • 11. Calculate the range, variance, and standard deviation of the Weekly Allowance of students in Binago School of Fisheries. Write your complete solutions and answers on a sheet of paper. Weekly allowance of Students in Binago School of Fisheries Weekly Allowance (in Pesos) Frequency Class Mark (X) fX (X - 𝑥) (X − 𝑥) 2 f(X − 𝑥) 2 500 – 549 2 450 – 499 3 400 – 449 1 350 – 399 3 300 – 349 4 250 – 299 14 200 – 249 12 150 – 199 21 100 - 149 10