This document defines key terms and concepts related to standard deviation and variance. It provides formulas for calculating range, deviation, variance, and standard deviation for both ungrouped and grouped data. Examples are given to demonstrate calculating these metrics from raw data sets and grouped data tables. Interpreting skewness is also discussed.

1. STANDARD
DEVIATION &
VARIANCE

2. 1) Range — Difference between max & min
Formula: Highest Value – Lowest Value
2) Deviation— Difference between entry &
mean
3) Variance—Sum of differences between
entries and mean, divided by population or
sample
Formula:
4) Standard Deviation —Square root of the
variance
Formula:
DEFINITION OF TERMS
2
Σ𝑓 𝑋 − 𝑥 ²
𝑛 − 1
Σ𝑓 𝑋 − 𝑥 ²
𝑛 − 1
Bessel’s
Correction
• the use of n − 1
instead of n in the
formula for the
sample variance and
sample standard
deviation, where n is
the number of
observations in a
sample.
• This method corrects
the bias in the
estimation of the
population variance.

3. UNGROUPED
DATA
3

4. Data: 17, 15, 23, 9, 9, 13
𝑠2
=
Σ 𝑋 − 𝑥 ²
𝑛 − 1
𝑠2 =
141.34
6 − 1
𝑠2
=
141.34
5
𝒔 𝟐 = 𝟐𝟐. 𝟓𝟔
𝑠 = 22.56
𝒔 = ±𝟒. 𝟕𝟓
UNGROUPED DATA
4
X X̅ X-X̅ (X-X̅ )²
17 14.33 2.67 7.13
15 14.33 0.67 0.45
23 14.33 8.67 75.17
9 14.33 -5.33 28.41
9 14.33 -5.33 28.41
13 14.33 -1.33 1.77
Σ 𝑋 − 𝑥 ²= 141.34

5. GROUPED
DATA
5

6. DATA
6
66 62 61
67 64 67
66 63 63
70 71 68
65 62 61
68 61 64
60 65 64
66 61 68
64 64 69
68 63 63
HEIGHT OF THE G11
STUDENTS AGES 15-19

7. 7
CLASS INTERVAL
Range HS-LS
Highest Score 71
Lowest Score 60
11
Range
]=
11 ≈ 2
√N 5
√N = √30 = 5.477225575 ≈ 5

8. GROUPED DATA
8
N= 30
Class
Interval
Frequency <Cf >Cf X f(X) X̅ X-X̅ (X-X̅ )² f(X-X̅ )²
60-61 5 5 30 60.5 302.5 64.83 -4.33 18.75 93.75
62-63 6 11 25 62.5 375 64.83 -2.33 5.43 32.58
64-65 7 18 19 64.5 451.5 64.83 -0.33 0.11 0.77
66-67 5 23 12 66.5 332.5 64.83 1.67 2.79 13.95
68-69 5 28 7 68.5 342.5 64.83 3.67 13.47 67.35
70-71 2 30 2 70.5 141 64.83 5.67 32.15 64.3
∑f(x)= 1945 ∑f(X-X̅ )²= 272.7

9. Mean
9
Median
𝑥 = 𝐿𝐿 +
𝑁
2
− 𝑐𝑓
𝑓
𝑖
𝑥 = 63.5 +
30
2
− 11
7
2
𝑥 = 63.5 +
15 − 11
7
2
𝑥 = 64.64
𝑥 =
Σ𝑓𝑥
𝑁
𝑥 =
1945
30
𝑥 = 64.83

10. Population

Σ𝑓 𝑋𝑚− 𝑥 ²
𝑛

272.7
30
 σ² = 9.09
VARIANCE
10
Sample

Σ𝑓 𝑋𝑚− 𝑥 ²
𝑛−1

272.7
30−1

272.7
29
 s² = 9.4

11. Population
= √ σ²
= √ 9.09
≈ ±3.01
STANDARD
DEVIATION
11
Sample
= √ s²
= √ 9.4
≈ ±3.07

12. 12
SKEWNESS

13. 13
SKEWNESS
SK =
3(Mean−median)
Standard Deviation
SK =
3(64.83−64.64)
3.07
SK = 0.19
Symmetric
Interpreting
1. If skewness is less than −1 or greater
than +1, the distribution is highly
skewed.
2. If skewness is between −1 and −½ or
between +½ and +1, the distribution
is moderately skewed.
3. If skewness is between −½ and +½,
the distribution is approximately
symmetric.

14. 14
SKEWNESS
0
1
2
3
4
5
6
7
60-61 62-63 64-65 66-67 68-69 70-71
Frequency
Height

15. -3 -2 -1 0 1 2 3
Example:
x = 65
μ = 64.83
σ = 3.07
Z SCORE
15
Solution:
=
65 − 64.83
3.07
= 0.55438
≈ 0.55

16. 16
FINDING THE PERCENTAGE UNDER THE
NORMAL DISTRIBUTION
Z~N(0,1)
P(Z) Q(Z) R(Z)
-3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3

17. 17
P(Z<0.55)
= 0.70884
= 70.84%
FINDING THE
PERCENTAGE UNDER
THE NORMAL
DISTRIBUTION
-3 -2 -1 0 1 2 3

18. 18
ACTIVITY
Class
Interval Frequency <Cf Xm fXm X̅ Xm-X̅ (Xm-X̅ )² f(Xm-X̅ )²
37-45 1 50 41 41 82.4 -41.4 1713.96 1713.96
46-54 0 49 50 0 82.4 -32.4 1049.76 0
55-63 2 49 59 118 82.4 -23.4 547.56 1095.12
64-72 6 47 68 408 82.4 -14.4 207.36 1244.16
73-81 14 41 77 1078 82.4 -5.4 29.16 408.24
82-90 13 27 86 1118 82.4 3.6 12.96 168.48
91-99 11 14 95 1045 82.4 12.6 158.76 1746.36
100-108 3 3 104 312 82.4 21.6 466.56 1399.68
Find the skewness
N= 50 ∑fxm= 4120 ∑f(Xm-X̅ )²= 7776
Clue: Find the mean, median and S.D. first
i= 9

19. Siberian Dogs
Mean: 48.5 lbs.
S.D.: 2.1 lbs.
Problem:
1.) Sketch the normal curve
2.) Find the percentage
- between 46.4 lbs. and 50.6 lbs.
Tip: Find the Z-scores first. Then, if you have a calculator, use Q(
and add both of them together.
19
ACTIVITY
-3 -2 -1 0 1 2 3

20. THANK YOU
20