Frequency Distribution and Relative Frequency Distribution of Students’ Marks

This document provides a frequency distribution and relative frequency distribution for 25 student’s marks with 5 classes. It also includes a frequency histogram, mean, standard deviation, range, and confidence interval.

STATISTICS (STAT-490) Pre MBA
Student Full Name:
Student ID:
CRN No:
Branch:

(i) A frequency distribution and a relative frequency distribution for 25 student’s marks
with 5 classes have been constructed below. The frequency distribution has been
constructed with Tally marks. The relative frequency table has been constructed and
provided in Table 2.
Table 1: Frequency Distribution of Students’ marks
Lower Upper Tally Mark FREQUENCY
45 54.8 ||| 3
54.8 64.6 |||| 5
64.6 74.4 |||| ||| 8
74.4 84.2 |||| 4
84.2 94 |||| 5
Class Boundaries
Table 2: Relative Frequency Distribution of Students’ marks
Lower Upper Midpoint FREQUENCY Rel.Freq
45 54.8 49.9 3 0.12
54.8 64.6 59.7 5 0.2
64.6 74.4 69.5 8 0.32
74.4 84.2 79.3 4 0.16
84.2 94 89.1 5 0.2
Class Boundaries
(ii) A frequency Histogram for 25 student’s marks with 5 classes has been created. The
horizontal scale has been taken as the marks of the students. The vertical scale
indicated the frequency of a particular class for the data.
Figure 1: Histogram of Students’ marks
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(iii) The mean and standard deviation for 25 student’s marks have been calculated as
below.
The mean mark of students is calculated using the formula
x
¿
=∑ xi
n
=¿ 70 + 80+ 86 + 46 + 56 + 66 +76 + 86 + 90 +70 + 50 + 45+94 + 65 +55 + 60 ¿
+ 90+ 80 + 70 + 71 + 72 + 62 + 64 + 76 +70 ¿ 25
=1750
25 =70
The standard deviation (S.D) of marks of students is calculated using the formula
s= √ ∑ ( xi −x
¿
)
2
n−1
“n-1” is taken instead of “n” in the denominator for calculating standard deviation for
the sample (Lee, In, and Lee, 2015, p. 220).
So, the calculated S.D =
√ ( 70−70 ) 2+ ( 80−70 ) 2+ ( 86−70 ) 2+. . .+ ( 64−70 ) 2+ ( 76−70 ) 2+ ( 70−70 ) 2
25−24
¿ √ 4428
24 =13. 583
(iv) Considering Mean = 70.7, SD = 13 and using Rule of thumb, we get
Minimum = Mean – 2*SD = 70.7 – 2*13 = 44.7
Maximum = Mean + 2*SD = 70.7 + 2*13 = 96.7
So, the range of student’s academic performance or score is 44.7 – 96.7 or
approximately 45 – 97 (Tipton, Hallberg, Hedges, & Chan, 2017, p. 472–505).
Hence, 95 would be within the interval 44.7 – 96.7, and would be considered “usual”.
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(v) According to the question, n=25 , x =70.7 and σ is known to be 13. Considering that
the sample is simple random, the 95% confidence interval for average(μ) marks is
calculated. At 95%, value of the z-statistic = 1.96 (Standardnormaltable 2016).
( x
¿
−z0 .95∗ σ
√ n , x
¿
+ z0 . 95∗ σ
√ n )= ( 70. 7−1. 96∗13
√ 25 , 70. 7+1 . 96∗13
√ 25 ) = ( 65 .604 ,75 . 796 )
Hence, with 95% confidence it is possible to say that student’s average academic score
(μ) is approximately estimated between 66 and 76.
References
Lee, D.K., In, J. and Lee, S., 2015. Standard deviation and standard error of the
mean. Korean journal of anesthesiology, 68(3), p.220.
Standardnormaltable 2016, The University of Arizona Department of Mathematics, retrieved
on February 19, 2019, from
<https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf>.
Tipton, E, Hallberg, K, Hedges, LV & Chan, W 2017, ‘Implications of small samples for
generalization: Adjustments and rules of thumb’, Evaluation review, vol. 41, no. 5, pp. 472–
505.
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Appendix
Table 3: Calculation for Standard Deviation in Excel
x (x-x-bar) (x-x-bar)^2
70 0 0
80 10 100
86 16 256
46 -24 576
56 -14 196
66 -4 16
76 6 36
86 16 256
90 20 400
70 0 0
50 -20 400
45 -25 625
94 24 576
65 -5 25
55 -15 225
60 -10 100
90 20 400
80 10 100
70 0 0
71 1 1
72 2 4
62 -8 64
64 -6 36
76 6 36
70 0 0
1750 0 4428
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