SUBJECT: MATHEMATICS
CLASS: SS 2
DATE:
TERM: 3rd TERM
REFERENCE BOOKS
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 |
|
|
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 |
= 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 + â1â1+â2C
Where Lm = lower class boundary of the modal class.
â1 = difference between the frequency of the modal class and the class before it.
â2 = 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, â1 = 35 – 30 = 5, â2 = 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 |
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 |
READING ASSIGNMENT
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 |
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