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02 Seatwork 3 - Skewness and Kurtosis, Assignments of Statistics

Finding the skewness and kurtosis.

Typology: Assignments

2023/2024

Uploaded on 09/09/2024

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BM2001
02 Seatwork 3 *Property of STI
Page 1of 3
Name: Curada, Maria Teresita P. Date: August 30, 2024
Subject: Statistical Analysis with Software Application Section: BSAIS 3
02 Seatwork 3
I. The following data are the weights (in pounds) of 40 female freshmen of STI College Ortigas
Cainta. Determine the coefficient of skewness and coefficient of kurtosis. (2 items x 5 points)
Class Interval
f
Class
Boundary
CM (x)
<CF
(x- x)
(x- x)4
f(x- x)4
119-130
4
118.5 - 130.5
124.5
4
22.20
242,891.27
971,565.06
131-142
11
130.5 - 142.5
136.5
15
10.20
10,824.32
119,067.54
143-154
15
142.5 154.5
148.5
30
1.80
10.50
157.46
155-166
7
154.5 166.5
160.5
37
13.80
36,267.39
253,871.76
167-178
3
166.5 178.5
172.5
40
25.80
443,076.61
1,329,229.83
Σf= 40
Σ(x- x)4=
773,070.09
Σf(x- x)4=
2,673,891.65
Mean = 146.7
Median = 146.5
Mode = 146.5
Standard deviation = 12.9
Skewness
��=3(
)
��=3(146.7146.5)
12.9 =0.6
12.9=�.������.��
where:=146.7
=146.5 =12.9
Kurtosis
�� =1
�=1
14
4
�� =1
40 2,673,891.65
12.94=0.025 2,673,891.65
27,692.29 =0.025 96.56 =�.��
where:=40 1()4=2,673,891.65 =12.9
pf2

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BM

02 Seatwork 3 _Property of STI_*

Page 1 of 3

Name: Curada, Maria Teresita P. Date: August 30, 2024

Subject: Statistical Analysis with Software Application Section: BSAIS 3

02 Seatwork 3

I. The following data are the weights (in pounds) of 40 female freshmen of STI College Ortigas

Cainta. Determine the coefficient of skewness and coefficient of kurtosis. (2 items x 5 points)

Class Interval f

Class

Boundary

CM (x) <CF (x- x̄ ) (x- x̄)

4

f (x- x̄)

Σ f = 40

Σ (x- x̄)

4

Σ f(x- x̄)

4

Mean = 146.

Median = 146.

Mode = 146.

Standard deviation = 12.

Skewness

𝐀

𝐀

where :

𝐀

= 146.7 𝐀 = 146.5 𝐀 = 12.

􃠀

𝐀

where :

𝐀

= 146.7 𝐀 = 146.5 𝐀 = 12.

Kurtosis

𵀀𴀀

𝐀=

𝐀

1

𝐀

4

4

𵀀𴀀

4

where :

𝐀 = 40 𝐀

1

(𝐀

𝐀

− 𝐀

)

4

= 2,673,891.65 𝐀 = 12.

BM

02 Seatwork 3 _Property of STI_*

Page 2 of 3

II. Give a brief interpretation on the data set. (5 points)

According to Karl Pearson, the proponent of Pearson’s Coefficient of Skewness, the data

provided to us is fairly symmetrical when plotted in a graph, if the coefficient falls between -0.

and 0.5. However, if the skewness is between 1 and -0.5 (negatively skewed) or between 0.

and 1 (positively skewed), the data is moderately skewed. Finally, if the skewness is less than -

1 (negatively skewed) or greater than 1(positively skewed), the data is highly skewed.

In the first coefficient of skewness, also called Pearson’s mode skewness, the value is 0.

which falls between -0.5 and 0.5. This means that the data is fairly symmetrical when plotted

in a graph. In the second coefficient of skewness, also called Pearson’s median skewness, the

value is 0.05 which still falls between -0.5 and 0.5. Thus, it also indicates that our data is

symmetrical.

As for the kurtosis coefficient, if the value is equal to 3, it represents that the data has a normal

distribution (called mesokurtic). If it is greater than 3, it means that the distribution has heavier

tails (called leptokurtic), indicating that more data is concentrated in the tails, therefore

suggesting outliers. A kurtosis value of less than 3, however, indicates a distribution with lighter

tails (called platykurtic), meaning less data is concentrated in the tails and more in the shoulders

(closer to the mean).

Since the kurtosis coefficient is 2.41 , which is less than 3, we can interpret that the distribution is

platykurtic , meaning it has lighter tails and is more flat or spread out compared to a normal

distribution. It also means that our data’s distribution (in a graph) is likely to have fewer extreme

values (outliers) than in a normal distribution.