WEEK TEN
TOPIC: STATISTICS: MEASURES OF CENTRAL TENDENCY
MEASUREMENT OF CENTRAL TENDENCY
- Mean, Median and Mode of ungrouped
- Mean, Median and Mode of grouped data
Measures of central tendency: This is a measure of how the data are centrally placed. The three commonest measures of position, depending on the information required are the arithmetic mean, median and the mode.
MEAN: It is most widely used measure and sometimes called the arithmetical averages. The mean of the number x1, x2, x3, x4 …………………xn is given by:
X = ∑x where ∑x is the sum of all items. n = number of items
n
When the data involves frequency; mean = ∑fx/∑f
Examples:
1. Calculate the mean of the numbers 15, 17, 19, 21, 23, 25, 27, 29.
Solution:
Mean (x) = 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 = 176 = 22
8 8
2. The table shows the number of suitcases possessed by a group of travelers.
| No of suitcases | 0 | 1 | 2 | 3 | 4 | 5 |
| Travelers | 2 | 7 | 7 | 2 | 3 | 9 |
Calculate the mean to the nearest whole number.
Solution:
Mean ( x ) = ∑ fx/∑ f = 84/30 = 2.8 = 3
| X | F | FX |
| 0 | 2 | 0 |
| 1 | 7 | 7 |
| 2 | 7 | 14 |
| 3 | 2 | 6 |
| 4 | 3 | 12 |
| 5 | 9 | 45 |
| Total | 30 | 84 |
EVALUATION
1. Calculate the mean of the numbers 37.5, 25.5, 30.5, 41.5, 52.5, 28.5.
2. Calculate the mean score of the scores represented in the table below.
| Scores | 10 | 12 | 14 | 16 | 18 |
| No of Students | 5 | 2 | 3 | 4 | 4 |
Mode:
The mode of a distribution is the value of the variable which occurs most often in the distribution. It is also possible for a distribution to have more than one mode, if there were more than one item having the highest frequency.
Example:
- Find the mode of the data 5, 4, 8, 9, 6, 8, 9, 3, 8. The mode is 8 (it appears 3 times more than others)
Median:
This is the middle value of a set of data, when arranged in ascending or descending order.
Example:
Find the median of these numbers: (1). 35, 28, 42, 28, 56, 70, 35 (2) 18, 20, 25, 30, 22, 25, 28, 15
Solution:
- Re – arranging the numbers: 70, 56, 42, [35] 35, 28, 28. The median is 35
- 15, 18, 20, [22, 25], 25, 28, 30. Median = 22 + 25 = 47 = 23.5
2 2
General Example:
The table below is the distribution of the test scored in a class:
| Scores | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Frequency | 1 | 1 | 5 | 3 | X | 0 | 6 | 2 | 3 | 4 |
If the mean score of the class is 6, find the (i) value of x (ii) median score (iii) modal score.
Solution:
| X | F | FX |
| 1 | 1 | 1 |
| 2 | 1 | 2 |
| 3 | 5 | 15 |
| 4 | 3 | 12 |
| 5 | X | 5x |
| 6 | 0 | 0 |
| 7 | 6 | 42 |
| 8 | 2 | 16 |
| 9 | 3 | 27 |
| 10 | 4 | 40 |
| Total | 25 + x | 155 + 5x |
- Mean = ∑fx/∑x
6 = 155 + 5x
25 + x
Cross multiplying: 6(25 + x) = 155 + 5x
150 + 6x = 155 + 5x
6x – 5x = 155 – 150
x = 5.
(ii)Median score: the median score lies between the 15th and 16th scores, hence: median = (5 + 7)/2 = 6.
(iii)Mode: 7
Evaluation:
Calculate the mode and median of the scores below; 2, 2, 1, 1, 0, 3, 3, 4, 4, 4, 5, 1, 2, 2.
MEAN, MEDIAN AND MODE OF GROUPED DATA
Mean: The arithmetic mean of grouped frequency distribution can be obtained using:
Class Mark Method:
X = 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 |
- Class Mark Method: X = = 3214/100 = 32. 14 = 32 matches per box (nearest whole no)
- 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 |
Histogram
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 |
15.5 18.5 21.5 24.5 27.5 30.5
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 + ∆1 C
∆1 + ∆2
Where Lm = lower class boundary of the modal class.
∆1 = difference between the frequency of the modal class and the class before it.
∆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, ∆1 = 35 – 30 = 5, ∆2 = 35 – 24 = 11, C = 3, Lm = 21.5
Mode = 21.5 + 5 x3
5 + 11
= 21.5 + (15/16) = 21.5 + 0.9375
= 22.44, approximately 22 yrs.
MEDIAN FROM INTERPOLATION FORMULA
Median = L1 + N/2 – cfm C
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 and 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 method.
| Shoe sizes | 30 – 34 | 35 – 39 | 40 – 44 | 45 – 49 | 50 – 54 |
| No of Men | 10 | 12 | 8 | 15 | 5 |
General evaluation:
1. 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 |
- State the modal class of the distribution and find the mode.
- Calculate the mean of the distribution.
2. The following table shows the distribution of marks obtained by a class.
| Marks | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| No of students | 1 | 1 | 3 | 4 | 4 | 12 | 7 | 3 | 3 | 2 |
Using this table, find the (1) median mark (2) modal mark (3) mean of the distribution.
Reading Assignment: Further Mathematics Project Book 1(New third edition), pg 328, Exercise18, No 15 -20
Weekend Assignment
| Marks | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 5 | x – 1 | x | 9 | 4 | 1 |
If the mean is 5, calculate the (a) value of x (b) mode (c) median of the distribution.
2. 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 |
- State the median class and calculate the median using interpolation method.
- Draw the histogram and use it to estimate the mode.
- Calculate the mean value using a working mean of 23.28mm.
- The table gives the frequency distribution of marks obtained by a group of students in a test.