MTH113 Statistics Assignment: Data Analysis and Interpretation

This document presents a comprehensive solution to a statistics assignment, covering various aspects of data analysis. The solution begins with data entry and sorting, followed by the calculation of range, mean, median, and mode. It then determines the class width, class limits, and class boundaries for a frequency distribution. The assignment further includes the creation of a frequency distribution table, a cumulative frequency table, and a stem and leaf diagram. Quartiles (Q1, Q3) and the interquartile range (IQR) are calculated, along with the 22nd percentile. Outliers are identified, and the standard deviation is computed using the sample population formula. Finally, a histogram visualizing the data is generated. This solution provides a step-by-step guide to solving statistical problems, demonstrating the application of key statistical concepts and techniques. Desklib offers a wealth of similar solved assignments and resources for students.

Running Head: STATISTICS ASSIGNMENT
Statistics Assignment
Name of the Student
Name of the University
Author Note

1STATISTICS ASSIGNMENT
Table of Contents
Answer a..........................................................................................................................................2
Answer b..........................................................................................................................................4
Answer c..........................................................................................................................................4
Answer d..........................................................................................................................................5
Answer e..........................................................................................................................................5
Answer f...........................................................................................................................................6
Answer g..........................................................................................................................................7
Answer h..........................................................................................................................................8
Answer i...........................................................................................................................................8
Answer j...........................................................................................................................................9
Answer k..........................................................................................................................................9
Answer l.........................................................................................................................................10
Answer m.......................................................................................................................................11

2STATISTICS ASSIGNMENT
Answer a
Step 1: Data Entry
Step 2: Sorting

3STATISTICS ASSIGNMENT
Step 3: Sorted data on Marks
The completely sorted data is given in the following table:
Stude
nt No.
Marks in
Maths
Stude
nt No.
Marks in
Maths
Stude
nt No.
Marks in
Maths
Stude
nt No.
Marks in
Maths
Stude
nt No.
Marks in
Maths
2 22 1 52 43 55 48 67 32 87
39 23 50 53 45 55 11 72 49 87
3 26 6 55 30 60 22 72 28 91
47 32 7 55 5 62 21 73 37 91
15 33 8 55 17 62 29 73 20 93
35 33 9 55 13 63 18 74 38 93
14 35 10 55 34 63 19 82 27 95
25 38 24 55 12 66 26 82 31 95

4STATISTICS ASSIGNMENT
4 39 41 55 23 66 33 82 40 99
46 42 42 55 16 67 36 82 44 99
Answer b
The range is given by the formula:
Range = Maximum Value – Minimum Value
= 99 – 22
= 77
Answer c
Mean = ∑ of all the Marks
Total Number of Students
= 3176
50 =63.52
Median = ( n
2 )
th
value+ ( n
2 +1 )
th
value
2
= ( 50
2 )th
value+( 50
2 +1)th
value
2
= ( 25 ) th value+ ( 26 ) th value
2
= 62+62
2
= 62

5STATISTICS ASSIGNMENT
Mode = The value which has occurred the most number of times
= 55.
Answer d
Range = 77
Number of classes = 8
Therefore, class width = 77
8 =9.6 10
Answer e
To find the class limits, the minimum limit is considered as 21 as the minimum marks obtained is
22.
The class width is 10.
Therefore, the lower limit of the 8 classes will be estimated as follows:
21
21 + 10 = 31
31 + 10 = 41
41 + 10 = 51
51 + 10 = 61
61 + 10 = 71
71 + 10 = 81

6STATISTICS ASSIGNMENT
81 + 10 = 91
91 + 10 = 101
The upper limit of the classes is calculated by subtracting 1 from the lower limit of the next
class. Thus, the upper limit of the classes is given as follows:
31 – 1 = 30
41 – 1 = 40
51 – 1 = 50
61 – 1 = 60
71 – 1 = 70
81 – 1 = 80
91 – 1 = 90
101 – 1 = 100
This will be the last class as no values are larger than 100 in the dataset. Therefore, the class
limits can be written as:
Class Limit
21 – 30
31 – 40
41 – 50
51 – 60
61 – 70
71 – 80
81 – 90
91 – 100

7STATISTICS ASSIGNMENT
Answer f
To find the class boundaries, the difference between the upper boundary of one class and the
lower boundary of the next class is divided by 2. The value thus obtained is added to the upper
class boundary and subtracted from the lower class boundary to obtain the class boundaries.
Upper limit of first class = 30
Lower limit of the second class = 31
Therefore, the class limits will be altered by = 31−30
2 =0.5
Thus, the class boundaries are given by:
Lower Boundary Upper Boundary Class Boundaries
21 – 0.5 = 20.5 30 + 0.5 = 30.5 20.5 – 30.5
31 – 0.5 = 30.5 40 + 0.5 = 40.5 30.5 – 40.5
41 – 0.5 = 40.5 50 + 0.5 = 50.5 40.5 – 50.5
51 – 0.5 = 50.5 60 + 0.5 = 60.5 50.5 – 60.5
61 – 0.5 = 60.5 70 + 0.5 = 70.5 60.5 – 70.5
71 – 0.5 = 70.5 80 + 0.5 = 80.5 70.5 – 80.5
81 – 0.5 = 80.5 90 + 0.5 = 90.5 80.5 – 90.5
91 – 0.5 = 90.5 100 + 0.5 = 100.5 90.5 – 100.5
Answer g
The frequency distribution and the cumulative frequency are given in the following table:
Class Boundaries Frequency Cumulative Frequency
20.5 – 30.5 3 3
30.5 – 40.5 6 9
40.5 – 50.5 1 10
50.5 – 60.5 13 23
60.5 – 70.5 8 31
70.5 – 80.5 5 36
80.5 – 90.5 6 42

8STATISTICS ASSIGNMENT
90.5 – 100.5 8 50
Grand Total 50
Answer h
The stem and leaf diagram of the data is given in the following table:
Stem and Leaf Display
Stem Leaves
2 2, 3, 6
3 2, 3, 3, 5, 8, 9
4 2
5 2, 3, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
6 0, 2, 2, 3, 3, 6, 6, 7, 7
7 2, 2, 3, 3, 4
8 2, 2, 2, 2, 7, 7
9 1, 1, 3, 3, 5, 5, 9, 9
Answer i
Quartiles can be estimated using the following formula:
Qi= ( ¿
4 )th
value
Here, N
4 =50
4 =12.5
Now, the 12.5th value lies between the 12th and the 13th value. The 12th value in the data is 53 and
the 13th value in the data is 55. Thus, the first quartile (Q1) is:
Q1= 53+55
2 =54
Again, 3 N
4 = 3 ×50
4 =37.5

9STATISTICS ASSIGNMENT
Now, the 37.5th value lies between the 37th and the 38th value. The 37th value in the data is 82 and
the 38th value in the data is 82. Thus, the third quartile (Q3) is:
Q3= 82+82
2 =82
The interquartile range (IQR) = Q3 – Q1 = 82 – 54 = 28
Answer j
Percentiles can be estimated using the following formula:
Pi= ( ¿
100 )th
value
Here, 22× N
100 =22 ×50
100 =11
Now, the 11th value is 52. Thus, the 22nd percentile is:
P22=52
Answer k
Outliers are the points that lies 1.5 IQR below the first quartile and 1.5 IQR above the
third quartile. Therefore,
Lower outlier range = Q1−1.5 × IQR=54−1.5× 28=12
Upper outlier range = Q3 +1.5 × IQR=82+ 1.5× 28=124
There are no values outside the ranges of the outliers in the given data on marks of
mathematics. Thus, there are no outliers to the data.

10STATISTICS ASSIGNMENT
Answer l
Class
Boundaries Mid Value (x) u = (x -
55.5)/10
Frequency
(f) uf (u^2)f
20.5 – 30.5 25.5 -3 3 -9 27
30.5 – 40.5 35.5 -2 6 -12 24
40.5 – 50.5 45.5 -1 1 -1 1
50.5 – 60.5 55.5 0 13 0 0
60.5 – 70.5 65.5 1 8 8 8
70.5 – 80.5 75.5 2 5 10 20
80.5 – 90.5 85.5 3 6 18 54
90.5 – 100.5 95.5 4 8 32 128
Grand Total 4 50 46 262
Therefore,
Standard Deviation=10 ×
√ ∑ u2 f
∑ f − (∑ uf
∑ f )2
¿ 10 × √ 262
50 − ( 46
50 )2
¿ 10 × √5.24− ( 0.92 )2
¿ 10 × √5.24−0.85
¿ 10 × √4.39
¿ 10 ×2.09 6
¿ 20.9 6

11STATISTICS ASSIGNMENT
Answer m
20.5 – 30.5 30.5 – 40.5 40.5 – 50.5 50.5 – 60.5 60.5 – 70.5 70.5 – 80.5 80.5 – 90.5 90.5 – 100.5
0
2
4
6
8
10
12
14
Histogram of Marks obtained in Maths
Class Boundaries
Frequencies