 Question 1

a)
 Class Frequency Upper Boundary Cumulative Process Cumulative Frequency 6 - 8 4 ≤ 8.5 4 4 9 - 11 6 ≤ 11.5 4 + 6 10 12 - 14 10 ≤ 14.5 10 + 10 20 15 - 17 3 ≤ 17.5 20 + 3 23 18 - 20 12 ≤ 20.5 23 + 12 35 Sum = 35
Table 1.1: “Less than or Equal” cumulative frequency polygon

Table 2.2: Cumulative Frequency of types “Less than or equal” polygon
 b)          Number of Classes, K= 1 + 3.3log (n) K= 1 + 3.3log (40) K= 6.28 6, or 6 to 7 classes Class width given = 10 Lower limit given = 50

 Number of classes Class Counting Tally Frequency (f) 1st class 50-59 IIII I 6 2nd class 60-69 IIII IIII IIII II 17 3rd class 70-79 IIII III 8 4th class 80-89 IIII IIII 9 Sum 40
Table 1.3: Frequency distribution table

 Class 50-59 60-69 70-79 80-89 Frequency (f) 6 17 8 9
Table 1.4: Frequency distribution table

Question 2

 Class (Given) Class mid-point (x) Frequency (f) (Given) 1.8 - 2.1 2 4.3 9.245 2.6 – 3.3 4 11.8 34.81 3.4 – 4.1 6 22.5 84.375 4.2 – 4.9 13 59.15 269.1325 5.0 – 5.7 8 42.8 228.98 5.8 – 6.5 3 18.45 113.4675 SUM 36 159 740.01
Table 2.1: Class mid-point, and calculation

Mean

Thus, the mean is 4.4

Mode
 Class Frequency (f) Class Boundaries Class width 4.2 - 4.9 13 4.15 - 4.95 (4.15 – 4.95) = 0.8

Calculation of class boundary  and

Class mode with the largest frequency is 4.2 – 4.9 (as per table 2.1).
Lower boundary =4.15
Class width, C = 0.8
The difference between the frequency of the class mode and the frequency of the class immediately before class mode, that is 13-6 = 7.
The difference between the frequency of the class mode and the frequency of the class immediately after class mode, that is 13-8 = 5
The mode formula
Thus, the mode is 4.62

Standard Deviation
Therefore, the standard deviation is 1.02
Question 3
a)         Frequency distribution table.
Number of Class,
K= 1 + 3.3log (n)
K= 1 + 3.3log (40)
K= 6.28 6 to 7 classes

Class width and class limits
Class width     = range / number of classes
= Largest number - smallest number / number of classes

 Number of Classes Class Counting Tally Frequency (f) 1st Class 4.8 - 7.1 IIII 5 2nd Class 7.2 – 9.5 IIII IIII 9 3rd Class 9.6– 11.9 IIII IIII IIII 15 4th Class 12 – 14.3 IIII 5 5th Class 14.4 – 16.7 III 3 6th Class 16.8 – 19.1 II 2 7th Class 19.2 – 21.5 I 1 SUM 40
Table 3.1: Frequency distribution table

 Class 4.8 - 7.1 7.2 – 9.5 9.6 – 11.9 12 – 14.3 14.4 – 16.7 16.8 – 19.1 19.2 – 21.5 Frequency (f) 5 9 15 5 3 2 1
Table 3.2: Frequency distribution table

b)
Mean
 Class Class mid-point (x) Frequency (f) (f) x (x) 4.8 - 7.1 5 29.75 7.2 – 9.5 9 75.15 9.6– 11.9 15 161.25 12 – 14.3 5 65.75 14.4 – 16.7 3 46.65 16.8 – 19.1 2 35.9 19.2 – 21.5 1 20.35 SUM 40 434.80
Table 3.3: Class mid-point calculation

Thus, the mean is 10.87

Mode
 Class Frequency (f) Class Boundaries Class width 9.6 – 11.9 15 9.55 – 11.95 11.95 – 9.55 = 2.4

Calculation of class boundary  and
Class mode with the largest frequency is 9.5 – 11.8 (as per table 3.3).
Lower boundary =9.55
Class width, C = 2.4
The difference between the frequency of the class mode and the frequency of the class immediately before class mode, that is 15-9 = 6.
The difference between the frequency of the class mode and the frequency of the class immediately after class mode, that is 15-5 = 10

The mode formula

Thus, the mode is 10.45

Median
 Class Frequency (f) Cumulative Frequency (F) Class Boundaries 4.8 - 7.1 5 0+5=5 7.2 – 9.5 9 5+9=14 9.6– 11.9 15 14+15=29 9.45-11.85 =2.4(width) 12 – 14.3 5 29+5=34 14.4 – 16.7 3 34+3=37 16.8 – 19.1 2 37+2=39 19.2 – 21.5 1 39+1=40 SUM 40
Table 3.4: cumulative frequency calculation

i)          The position of the median:
ii)         The class median with a cumulative frequency (f) > 20.5 is the class median
9.6 – 11.9.
iii)        The median  is:

Thus, the median is 10.59
Question 4
a)                  The value of :
Thus, the value of x = 4

Question b,c,d and e are based on the table 4.1 below
 Class Number of Students (f) Cumulative Frequency (F) Class boundaries Class width 40 - 44 2 0+2=2 45 - 49 2+4=6 50 - 54 7 6+7=13 49.5 – 54.5 5 55 - 59 13+16=29 54.5 - 59.5 5 60 - 64 29+20=49 59.5 – 64.5 5 65 - 69 2 49+2=51 70 - 74 1 51+1=52 SUM 52
Table 4.1: Coursework Statistic Marks

b)         First Quartile
Step 1

Step 2
 Quartile 1
Cumulative frequency (f) of class quartile with F > 13.25, that is F = 29
 55-59 16 13+16=29 54.5 - 59.5 5

Step 3
Therefore, the first quartile is 54.58

c)         Median
i)          The position of the median:
ii)         The class median with a cumulative frequency (f) > 26.5 is the class median
55 – 59.
iii)        The median  is:

Thus, the median is 58.72.

d)         Third Quartile
Step 1

Step 2
 Quartile 3
Cumulative frequency (f) of class quartile with F > 75.75, that is F = 49
 60-64 20 29+20=49 59.5-64.5 5

Step 3
Therefore, the third quartile is 62.19

d)         Inter-Quartile Range

IQR ;

Therefore, the IQR is 7.61