Lesson Notes By Weeks and Term - Senior Secondary School 2

MEAN, MEDIAN AND MODE OF GROUPED DATA

SUBJECT: MATHEMATICS

CLASS:  SS 2

DATE:

TERM: 3rd TERM

REFERENCE BOOKS

• New General Mathematics SSS2 by M.F. Macrae etal.
• Essential Mathematics SSS2 by A.J.S. Oluwasanmi.
• Exam Focus Mathematics.

WEEK SEVEN

TOPIC: MEAN, MEDIAN AND MODE OF GROUPED DATA

MEAN: The arithmetic mean of grouped frequency distribution can be obtained using:

Class Mark Method:

X  =  fx/∑f    where x is the midpoint of the class interval.

Assumed Mean Method: It is also called working mean method.    X  =  A + (∑ Fd/∑f)

Where, d = x – A,   x = class mark and A = assumed mean.

EXAMPLE: The numbers of matches in 100 boxes are counted and the results are shown in the table below:

 Number of matches 25    -   28 29    -    32 33   -    36 37      -     40 Number of boxes 18 34 37 11

Calculate the mean (i) using class mark    (ii) assumed mean method given that the assumed mean is 30.5.

Solution:

 Class interval F X FX d = x - A Fd 25     -     28 18 26.5 477 4 72 29     -     32 34 30.5 1037 0 0 33     -     36 37 34.5 1276.5 4 148 37     -     40 11 38.5 423.5 8 88 Total 100 3214 164

1. Class Mark Method: X  =  fx/∑f   =  3214/100   = 32. 14 = 32 matches per box (nearest whole no)
2. Assumed Mean Method: X  =  A + (∑ Fd/∑f)

= 30. 5 + (164/100) =30.5 + 1.64

= 32.14 = 32 matches per box (nearest whole number)

EVALUATION:

Calculate the mean shoe sizes of the number of shoes represented in the table below using (i) class mark   (ii) assumed mean method given that the assumed mean is 42.

 Shoe sizes 30    -      34 35      -    39 40     -     44 45      -    49 50      -   54 No of Men 10 12 8 15 5

MODE

The mode of a grouped frequency distribution can be determined geometrically and by interpolation method.

Mode from Histogram: The highest bar is the modal class and the mode can be determined by drawing a straight line from the right top corner of the bar to the right top corner of the adjacent bar on the left. Draw another line from the left top corner to the bar of the modal class to the left top corner of the adjacent bar on the right.

Example:

The table gives the distribution of ages of students in an institution.

 Ages(year) 16    -      18 19      -    21 22     -     24 25      -    27 28      -   30 No of Students 18 30 35 24 13

Draw a histogram and use your histogram to estimate the mode to the nearest whole number.

Solution:

 Class Interval (Ages) F Class Boundary 16      -     18 18 15.5    -   18. 5 19     -      21 30 18.5    -   21.5 22     -      24 35 21.5    -   24.5 25     -      27 24 24.5    -  27.5 28     -      30 13 27.5   -   30.5

35

30

25

20

15

10

5

0

15.5     18.5    21.5     24.5    27.5    30.5                                  Histogram

Modal class = 22   -    24

Mode = 21.5 + 0.9 = 22.4, approximately 22 yrs.

MODE FROM INTERPOLATION: The mode can be obtained using the formula.

Mode = Lm + â1â1+â2C

Where Lm = lower class boundary of the modal class.

â= difference between the frequency of the modal class and the class before it.

â2  = difference between the frequency of the modal class and the class after it.

C   = class width of the modal class.

Example: Using the table given in the example above:

Modal class = 22 – 24,   â1 = 35 – 30 = 5,  â2 = 35 – 24 = 11,   C = 3,   Lm = 21.5

Mode = 21.5   +     5            3

5 + 11

= 21.5 + (15/16)   = 21.5 + 0.9375

= 22.44, approximately 22 yrs.

MEDIAN OF GROUPED DATA: The median of grouped data can be determined from a cumulative frequency curve and from the interpolation formula.

Median from Cumulative Frequency Curve: The cumulative frequency curve can be used to determine the median.

EXAMPLE: The table below shows the masses of 50 students in a secondary school

 Masses (kg) 10    -   14 15    -   19 20  -     24 25    -   29 30  -  34 35 -  39 40    -     44 Frequency 3 7 9 5 11 6 9
1. Prepare a cumulative frequency table for the data.
2. Draw the ogive and use your graph to find the median.

Solution:

50                                       *

45

40                                           *

35                  *

30

25          *

20                           *

15

10                     *

5             *

0

 Masses(kg) Frequency Cumulative Frequency Upper Class Boundary 10 – 14 3 3 < 14.5 15 – 19 7 10 <19.5 20 – 24 9 19 <24.5 25 – 29 5 24 < 29.5 30 – 34 11 35 < 34.5 35 – 39 6 41 < 39.5 40 – 44 9 50 < 44.5

14.5   19.5   24.5    29.5  34.5  39.5  44.5     Upper Class Boundary

Cumulative Frequency Curve Showing the Masses of 50 Students.

To find the median, find (N/2) and check the table on the curve.

Therefore, N/2  = 50/2   = 25th

Check 25th on the cumulative frequency and trace to the upper class boundary.

Median = 29.5 + 0.5 = 30kg

MEDIAN FROM INTERPOLATION FORMULA

Median = L1 +   N/2 – cfmC

fm

Where, L1 = lower class boundary of the median class.

Cfm = cumulative frequency of the class before the median class.

Fm = frequency of the median class.

C   = class width of the median class.

N   = Total frequency

The median class: 30 – 34, L1 = 29.5, cfm = 24,   fm = 11,   C = 5

Median = 29.5 +  25 - 24   x 5

11

= 29.5 + 5     = 30kg

EVALUATION: Calculate the modal shoe sizes and median of the number of shoes represented in the table below using interpolation and graphical method.

 Shoe sizes 30    -      34 35      -    39 40     -     44 45      -    49 50      -   54 No of Men 10 12 8 15 5

GENERAL EVALUATION:

The table below gives the distribution of masses (kg) of 40 people

 Masses (kg) 1 - 5 6 – 10 11 -15 16    -   20 21     - 25 26    - 30 31     - 35 36    -   40 Frequency 9 20 32 42 35 22 15 5
1. State the modal class of the distribution and find the mode.
2.  Draw a cumulative frequency curve to illustrate the distribution.
3. Use the curve in ‘2’ to estimate the median.
4. Calculate the mean of the distribution.

New General Mathematics SSS2,page 160,  exercise14a.

WEEKEND ASSIGNMENT

The table gives the frequency distribution of a random sample of 250 steel bolts according to their head diameter, measured to the nearest 0.01mm.

 Diameter (mm) 23.06 – 23.10 23.11 – 23.15 23.16 – 23.20 23.21 – 23.25 23.26-23.30 23.31 – 23.35 23.36-23.40 23.41-23.45 23.46-23.50 No of bolts 10 20 28 36 52 38 32 21 13
1. State the median class and calculate the median using interpolation method.
2. Draw the histogram and use it to estimate the mode.
3. Calculate the mean value using a working mean of 23.28mm.