Lesson Notes By Weeks and Term - Senior Secondary School 1

Calculation of Range, Median and Mode of Grouped data

SUBJECT: MATHEMATICS

CLASS:  SS 1

DATE:

TERM: 3rd TERM

REFERENCE BOOKS

  • New General Mathematics SSS 1 by M.F. Macrae et al 
  • Essential Mathematics SS 1

 

 
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                    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

  1. Which of the following is the same as the arithmetic mean of a distribution?A.Mean deviation B. average C. Ordinary mean D. Percentage
  2. A bundle of tally consists of ____________ strokes?A.12 B. 10 C. 5 D. 4
  3. Frequency is defined as the…………………………..A.The number of times a variable occur in a distribution          B. The number of bundles in a cell of tallies     C. The highest occurrence scores   D. The average score
  4. The range of the distribution: -2,3,3,1,1.7,2.4 and 2.6 is _____ ?    A. 4  B. 0.6 C. 4.6 D. 3.5
  5. `Find the average age of the following distribution:1.23,2.32,1.17,2,3.11,2.11and2.12

 

THEORY

  1. A group of students were asked to state their year of birth,the results are as  follows

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. prepare a frequency table for this data                
  2. which year of birth has this highest frequency
  3. what fraction and percentage of the student were born in 1990 and above  

 

  1. The height in meters of student in sss1 class in a certain secondary school were given as follows

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

  1. Prepare a frequency distribution table for this data
  2. How many student are in sss1?
  3. What is the different between the highest and lowest height in cm?
  4. How many student are more than 1.3 m tall?
  5. What percentage of the student are 1.3 m tall and less?
  6. State whether the data is discrete or continuons














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