Calculation of Range, Median and Mode of Grouped data
SUBJECT: MATHEMATICS
CLASS: SS 1
DATE:
TERM: 3rd TERM
REFERENCE BOOKS
WEEK NINE
TOPIC: Calculation of Range, Median and Mode of Grouped data
RANGE: This is defined as the difference between the HIGHEST variable and the LEAST variable.
Example: Find the range of the following distribution: 2.2, 2.5, 2.2, 1.6, 1.8, 2.7,and 1.4
Solution: Range= Highest score – Lowest score
Highest score = 2.8
Least score = 1.4
Range = 2.8 – 1.4 = 1.4
The above example is ungrouped data; therefore, the range is as simple as that.
To find the range from Grouped, just identify the highest (Upper) class interval and the Least (Lower) Class interval and find the difference.
Example 1: find the range of the distribution:
1-10 11-20 21-30 31- 40 and 41- 50
Highest = 50
Least = 1
Range = 50 – 1 = 49
THE MEAN: This is also known as Arithmetic mean, it is denoted with the symbol X.Simply put, arithmetic mean is also known as average.
For simple data, Such as: EXAMPLE (1) 2.2, 2.5, 2.2, 1.6, 1.8, 2.7, and 1.4, to calculate the arithmetic mean,the required formula is the same as that of the average: e g
MEAN = SUM OF THE ALL VARIABLES/SCORES
NUMBER VARIABLES/SCORES
2.2 + 2.5 + 2.2 + 1.6 + 1.8 + 2.7 + 1.4 = 14.4 = 2.06
7 7
The basic formular for the calculation of the arithmetic mean is given below:
X = ∑Fx where,∑ (Sigma) means summation.
∑F
Hence,Mean (X) = Sum of the product of the frequency and scores
Sum of the frequencies
EXAMPLE 2: The table below gives the scores of a group of students in a mathematics test
SCORES | 2 | 3 | 4 | 5 | 6 | 7 |
Number of Students | 2 | 4 | 7 | 2 | 3 | 2 |
Calculate the mean mark of the distribution:
Solution (Method,1)
Mean = ∑Fx= ( 2X 2) + (3 X 4) + (4 X 7) + ( 6X 3) + ( 7X2)
∑F 2 + 4 + 7 + 2 + 3 + 2
= 4 + 12 + 28 + 18 + 14
20
= 86 = 4.3
20
(Method 2): A simple frequency distribution may be constructed
SCORES | FREQUENCY(f) | Fx | |
2 | 2 | 4 | |
3 | 4 | 12 | |
4 | 7 | 28 | |
5 | 2 | 10 | |
6 | 3 | 18 | |
7 | 2 | 14 | |
∑f = 20 | ∑fx= 86 |
∑fx= 86 and ∑ = 20
therefore,Mean = 86 = 4.3
20
ARITHMETIC MEAN FROM GROUPED DATA:
To calculate the arithmetic mean from grouped data, a frequency table is necessary, only the Class intervals, frequencies, class marks(Mid Mark) and fx column is required.
EXAMPLE 3:The distributions of the waiting time for some students in a school is given as follows:
Waiting Time (minuetes) Number of customers
1.5 – 1.9 3
2.0 – 2.4 10
2.5 – 2.9 18
3.0 – 3.4 10
3.5 – 3.9 7
4.0 – 4.4 2
Calculate the mean time of the distribution:
Solution: Prepare a simple frequency distribution table for a grouped data:
Time intervals (Minuetes) | Mid Time (x) | No of Students Frequencies | Fx |
1.5 – 1.9 | 1.7 | 3 | 5.1 |
2.0 – 2.4 | 2.2 | 10 | 22.0 |
2.5 – 2.9 | 2.7 | 18 | 48.6 |
3.0 – 3.4 | 3.2 | 10 | 32.0 |
3.5 – 3.9 | 3.7 | 7 | 25.9 |
4.0 – 4.4 | 4.2 | 2 | 8.4 |
∑ f = 50 | ∑fx=142.0 |
Mean Time (Average Time) = ∑fx= 142.0 = 2.8 minutes
∑f 50
THE MODE
The mode is the variable or score with the highest frequency. The variable with the highest occurrence or which appears most in an event is known as the MODE.
EXAMPLE: Determine the modal mark in the distribution table:
Marks | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Frequency | 5 | 3 | 2 | 6 | 5 | 1 | 3 |
Solution: Modal mark = 7 with the frequency of 6.
It is possible to record more than one variable as the modal mark.When only one number appears most (as mode) it is said UNIMODAL.When two numbers appears as the mode it is said to be BIMODAL and when more than two numbers appear as mode it is said to be MULTIMODAL.
THE MEDIAN
Median is the number(s) which appears at the middle.It is possible for two numbers to appear at the middle,especially when the total variable is even number,in such a case,the average of the two mid numbers,is calculated as the MEDIAN.it must be noted that before the median is picked or calculated,the variables or scores must be arranged in an order of magnitude.i.e,ascending or Descending Order of Magnitude.
EXAMPLE: Calculate the median of the distribution:
2, 6, 4, 5, 5, 8, 8, 6, 6, 5, 9, 9, 2, 7, 4, 6, 3, 5, 6, 2, 7, 2, 9, 8, 10,6
Steps in the variables in an order of magnitude
:2,2,2,2,3,4,4,5,5,5,5,6,6,6,6,6,6,7,7,8,8,8,9,9,9,10 calculating MEDIAN from ungrouped (even) variables
STEP (i) Rearrange
STEP(ii),Divide total number by 2. i.e,26/2=13.
STEP (iii) Count 13 numbers from both left and right
STEP (iv) subtract one from each, result is 12.Hence 12 numbers are then counted from both left and right as shown below: 2,2,2,2,3,4,4,5,5,5,5,6, 6,6 ,6,6,6,7,7,8,8,8,9,9,9,10
From the above, two numbers are at the centre (6, 6) therefore the average of these numbers is the
median= 6 + 6 = 12 = 6. Therefore,median = 6
2 2
EXAMPLE: 2
Find the median of the scores below:
2.0, 1.8, 3.9, 4.5, 2.6, 3.7, 5.0, 2.1 and 3.3
Solution:
Rearranging the scores: 1.8, 2.0 ,2.1, 2.6, 3.3, 3.7, 3.9, 4.5, 5.0
There are nine scores in all; 9/2= 4.5
Counting four numbers from both left and right 1.8, 2.0 ,2.1, 2.6, 3.3, 3.7, 3.9, 4.5, 5.0
MEDIAN = 3.3
MEDIAN FROM TABLES:
EXAMPLE 3: The table shows the marks scored by SSS 1 students in a mathematics test
MARK | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
FREQUENCY | 5 | 3 | 2 | 6 | 5 | 1 | 3 |
Find the median
Make a table as follows:
Marks ( x ) Frequency (f)
4 5
5 3
6 2
7 6
8 5
9 1
10 3
Position of Median = ∑f + 1 = 25 + 1 = 26 = 13
2 2 2
Counting down the frequency column as shown on the above table,the position of the median (i.e,13th position) occurs opposite 7.
Thus the median mark = 7
EVALUATION:
The table gives the frequency distribution of marks obtained by a group of students in a test
Marks | 3 | 4 | 5 | 6 | 7 | 8 |
Frequency | 5 | X – 1 | x | 9 | 4 | 1 |
If the mean mark is 5 (a) Calculate the value of x
(b)Find the (i) mode (ii) Median (iii) Range of the distribution
READING ASSIGNMENT
Essential Mathematics for Senior Secondary Schools 1 pg 336 – 351
WEEKEND ASSIGNMENTS
OBJECTIVES
THEORY
1990 1992 1990 1989 1991 1990
1990 1988 1990 1989 1989 1991
1992 1992 1990 1989 1988 1990
1991 1991 1990 1988 1992 1991
1990 1990 1992 1991
1.3 1.3 1.2 1.4 1.2 1.5 1.5 1.4 1.3 1.6
1.6 1.5 1.3 1.6 1.3 1.4 1.5 1.3 1.2 1.1
1.3 1.2 1.5 1.5 1.4 1.3 1.2 1.4 1.6 1.5
1.4 1.5 1.2 1.1 1.6 1.5 1.5 1.5 1.5 1.4
1.2 1.3 1.4 1.5 1.4 1.5 1.5 1.4 1.3 1.2
1.5 1.5
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