MEASURES OF DISPERSION
WEEK 4
SUBJECT: FURTHER MATHEMATICS
CLASS: S.S 1 THIRD TERM
TOPIC: MEASURES OF DISPERSION
CONTENT:
- Range
- Inter-quartiles
- Mean deviation
- Standard deviation
- Variance
- Coefficient of variation
Range
Measures of Dispersion
The measure of dispersion (also called measure of variation) is concerned with the degree of spread of the numerical value of a distribution.
The Range: This is the difference between the maximum and minimum values in the data.
Examples 1:
Find the range of the data 6, 6, 7, 9, 11, 13, 16, 21 and 32
Solution: The maximum item is 32
The minimum item is 6
∴ Range = 32 – 6 = 26
Examples 2:
The following shows the distribution of mails obtained by 20 students in a test:
58 55 52 56 64 71 64 53 54 67
66 68 78 52 65 57 61 65 59 52
Find the range.
Solution
The largest value in the distribution = 78
The smallest value in the distribution = 52
Range = 78 – 52
= 26
Inter-quartiles
The difference between the values of the third and first quartiles is called the inter-quartile range. In equation form, we state this as:
Inter-quartile range = Q_{3} – Q_{1}
Inter-quartile range is a measure of spread.
Quartile deviation
One-half the interquartile rang is another measure of spread and is called quartile deviation. Another name for quartile deviation is semi-interquartilerange. In equation form, we state this as:
Quartile deviation (Semi-interquartile range)=(Q_{3} – Q_{1})
Recall that Q_{1,} Q_{2}and Q_{3 }can be obtained from a cumulative frequency curve.
Compute to find their position on the cumulative frequency (CF) axis using the following formulae:
(a) For lower quartile or first quartile () we use
(b) For median quartile or second quartile (), we use
(c) For upper quartile or third quartile (), we use (Total frequency or last CF)
Cumulative frequency
Upper class boundaries
STEP 2: Locate the point on the cumulative frequency axis and draw a horizontal line from this point to intersect the Ogive.
STEP 3: At the point it intersect the Ogive, draw a line parallel to the cumulative frequency axis to intersect the horizontal axis.
STEP 4: Read the value of the desired quartile at the point of intersection of the vertical line and the horizontal axis.
Inter-quartile range =
Semi inter-quartile range =
Inter –Percentile Range:- This is the difference between the 90^{th} percentile and 10^{th} percentile. Usually denoted by 10 90 percentile Range = P_{90} P_{10.}
Note: Interpolation formula on lower and upper quartiles should be studied/taught by student/teacher exhaustively.
CLASS ACTIVITY
Choose the correct answer from the options:
- The highest mark in a mathematics class is 80% and the lowest mark in the class is 25%. What is the range of the marks in the class?
- 105% (b) 52.5% (c) 55% (d) 6.85%
- In a distribution, the lower quartile is found to be 41.5 and the upper quartile is 55.2. What is the interquartile range of the distribution?
- 41.5 (b) 13.7 (c)48.35 (d) 6.85
Mean deviation
Also called average deviation of a set of numbers x_{1 },x_{2}…….xn is defined by
Where is the arithmetic average of the numbers and is the absolute value of the deviation of from .n is n is the number of terms.
Example 1: Find the mean deviation of the numbers 3,8,4,5,1,9
Arithmetic mean,
=
= 5
M.D. |3-5|+|8-5|+|4-5|+|5-5|+|1-5|+|9-5|)
MEAN DEVIATION OF A GROUP DATA.
Example2:Find the mean deviation of the distribution:
4,3,5,2,6,3,7,5,1,5,4,6,5,1,8.
x | F | f(x) | |x – | F|x – | |
1 | 2 | 2 | ||
2 | 1 | 2 | ||
3 | 2 | 6 | ||
4 | 2 | 8 | ||
5 | 4 | 20 | ||
6 | 2 | 12 | ||
7 | 1 | 7 | ||
8 | 1 | 8 | ||
15 | 65 | 16 |
STANDARD DEVIATION:
The standard deviation of a distribution x_{i},x_{2}…….x_{n} denoted by S is defined by the formula;
For grouped data
Example 1:
Find the standard deviation of the set of number 8,9,4,3,1,5
Solution:
X | ||
8 | 3 | 9 |
9 | 4 | 16 |
4 | -1 | 1 |
3 | -2 | 4 |
1 | -4 | 16 |
5 | 0 | 0 |
Total | 46 |
Using alternative method
Solution:
X | X |
8 | 64 |
9 | 81 |
4 | 16 |
3 | 9 |
1 | 1 |
5 | 25 |
30 | 196 |
= 2.77
Example 2:
The table below shows the masses of 200 students in a school. Calculate the standard deviation.
Mass in kg | F |
0-9 | 24 |
10-19 | 68 |
20-29 | 63 |
30-39 | 18 |
40-49 | 10 |
50-59 | 8 |
60-69 | 5 |
70-79 | 5 |
Solution :
Mass in kg | F | c- | f( | |
0-9 | 24 | 4.5 | -40 | -960 |
10-19 | 68 | 14.5 | -30 | -2040 |
20-29 | 63 | 24.5 | -20 | -1260 |
30-39 | 18 | 34.5 | -10 | -180 |
40-49 | 10 | 44.5 | 0 | 0 |
50-59 | 8 | 54.5 | 10 | 80 |
60-69 | 5 | 64.5 | 20 | 100 |
70-79 | 4 | 74.5 | 30 | 120 |
200 | -4140 |
Let =44.5
= 44.5 +
X | f | |||
4.5 | -19.3 | 24 | 372.49 | 8939.76 |
14.5 | -9.3 | 68 | 86.49 | 5881.76 |
24.5 | 0.7 | 63 | 0.49 | 30.87 |
34.5 | 10.7 | 18 | 114.49 | 2060.82 |
44.5 | 20.7 | 10 | 428.49 | 4284.9 |
54.5 | 30.7 | 8 | 942.49 | 7539.92 |
64.5 | 40.7 | 5 | 1656.49 | 8282.45 |
74.5 | 50.7 | 4 | 2560.49 | 10241.96 |
200 | 47261.00 |
.
Using the alternative method
or
X | F | Fx^{2} | F(x) | |
4.5 | 24 | 20.25 | 486 | 108 |
14.5 | 68 | 210.25 | 14297 | 987 |
24.5 | 63 | 600.25 | 37815.75 | 1543.5 |
34.5 | 18 | 1990.25 | 21424.5 | 621 |
44.5 | 10 | 1980.25 | 19802.5 | 445 |
54.5 | 8 | 2970.25 | 23762 | 436 |
64.5 | 5 | 4160.25 | 20801.25 | 322.5 |
74.5 | 4 | 5550.25 | 22201 | 298 |
; ; = 22657600;
⇒
Variance
This is also called the mean squared deviation. Variance is therefore the square of the standard deviation. This can be computed using the formula,
This means that the variance of the example above is (15.38)^{2}.
Example 1:
The table gives the distribution of marks obtained by a group of 10 students in a test. Calculate the variance of the distribution.
Marks | 1 | 3 | 5 | 7 | 9 |
No. of students | 4 | 2 | 2 | 1 | 1 |
Solution
x | x^{2} | f | fx | f x^{2} |
1 | 1 | 4 | 4 | 4 |
3 | 9 | 2 | 6 | 18 |
5 | 25 | 2 | 10 | 50 |
7 | 49 | 1 | 7 | 49 |
9 | 81 | 1 | 9 | 81 |
Ʃf = 10, Ʃfx = 36, Ʃfx^{2}= 202,
Let the variance be s^{2}
=
= 20.2 – 3.6^{2}
= 20.2 – 12.96
= 7.24
Example 2:
Calculate the variance of the numbers 19, 21, 27 and37
Solution
X | |
19 | 361 |
21 | 441 |
27 | 729 |
37 | 1369 |
,
Let the variance be then
=
= 725 – 676
= 49
CLASS ACTIVITY
- Marks of some students is tabulated below :
X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
F | 1 | 2 | 2 | 0 | 5 | 6 | 8 | 5 | 3 | 2 | 1 |
Use this to find:
- Range
- Mean deviation
- Standard deviation of the distribution
- Use this set of numbers 5, 3, 6, 8, 3 to answer questions (iii)—–(v)
- The standard deviation S.D is
(a) 1 (b) 1.9 (c) 1.6 (d) 0.8
- The mean deviation is
- 1.6 (b) 1.0 (c) 1.6 (d) 0.8
- The variance of the distribution is
- 1 (b) 0.64 (c) 2.56 (d) 3.6
Coefficient of variation
So far we have studied the various measures of dispersion of data in distributions. We can see that the most reliable of them all is the standard deviation. The standard deviation though reliable yet it has its own deficiency in that if we which one is most widely spread, the data that has bigger values will have a bigger standard deviation than the other one with lesser values. Any decision taking from such comparison will be misleading. In order to overcome this, we then find coefficient of variation. This one is independent of magnitude for better comparison.
How then do we find coefficient of variation?
Coefficient of variation= X 100%
Coefficient of standard deviation =
Example 1: given two sets of the following numbers, find which one is more widely dispersed.
A = 2, 3, 1, 4, 5, 6, 7, 25, 9, 8
B = 32, 45, 40, 38, 47, 50, 43, 36, 44, 35
Solution
Let _{A }represents the mean of A
Let _{B }represents the mean of A
Let the standard deviation for these sets be represented by S_{A}, S_{B}.
_{A }= = 7
_{B }= = 41
S_{A} =
= 6.48
S_{B} =
S_{B} = = 5.46
The coefficient of standard deviation of A =
=
The coefficient of standard deviation of B =
=
The coefficient of variation is multiplied by 100%. A = 92.6% and B = 13.3%
From the calculation Data A is more widely dispersed than data B.
Let us get a new set of B by multiplying by 5 throughout and call it data C.
= 160 + 225 + 200 + 190 + 235 + 250 + 215 + 180 + 220 + 175 ÷ 10 = 205
S_{c} =
= 2050 + 400 + 25 + 225 + 900 + 2050 + 100 + 625 + 225 + 900
= 7450 ÷ 205
= 27.29
The coefficient of standard deviation of C = 0.133
C .V = 13.3%
From the example above coefficient of variation does not depend on the large numbers. This makes if more reliable.
CLASS ACTIVITY
- An epidemic affected a village and the following was the record
Age | 10 – 12 | 12 – 14 | 14 – 16 | 16 – 18 | 18 – 20 | 20 – 22 | 22 -24 |
No Affected | 5 | 12 | 20 | 26 | 12 | 1 | 4 |
- Which age – group was most affected
- Calculate the standard deviation
- Calculate also the coefficient variation (C.V)
- Find the Mean, Standard deviations and coefficient of variations of the following distribution
X | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
F | 1 | 3 | 4 | 9 | 8 | 3 | 2 |
- What is the Mean, Standard deviations and coefficient of variations of the following distribution?
X | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
F | 1 | 3 | 4 | 9 | 8 | 3 | 2 |
What can you notice about questions 2 and 3?
- The following values of a quantity x were obtained experimentally.
X | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
F | 2 | 1 | 4 | 7 | 10 | 12 | 8 | 5 | 1 |
Calculate the mean and the standard deviation. Hence compute the coefficient of standard deviation and coefficient of variation.
- Compute (i) the variance (ii) the standard deviation of the following test scores.
Marks | 10 | 12 | 13 | 14 | 15 | 16 |
Frequency | 18 | 4 | 6 | 12 | 6 | 4 |
PRACTICE QUESTIONS
- The table below shows the distribution of the waiting times for some customers in a certain petrol station.
Waiting time (in min) 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
(a) Write down the class boundaries of the distribution.
(b) Construct a cumulative frequency curve for the data.
(c) Using your graph, estimate:
(i) the interquartile range of the distribution.
(ii) the proportion of customers who would have waited for more than 3 minutes.
- The frequency distribution of the weight of 100 participants in a high jump competition is as shown below:
Weight (kg) | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 |
Number of participants | 10 | 18 | 22 | 25 | 16 | 9 |
(a) Construct the cumulative frequency table
(b) Draw the cumulative frequency curve
(c) From the curve, estimate the:
(i) median
(ii) semi-interquartile range
(iii) probability that a participant chosen at random weighs at least 60kg.
- The following table shows the distribution of marks obtained by some students in an examination.
Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40 -49 | 50 – 59 | 60-69 | 70-79 | 80-89 | 90-99 |
No. of students | 50 | 50 | 40 | 60 | 100 | 100 | 50 | 25 | 15 | 10 |
- Construct a cumulative frequency table for the distribution.
- Draw an ogive for the distribution.
- Use your graph in (b) to determine the;
- Semi-inter quartile range;
- Number of students who failed, if the pass mark for the examination is 37;
- Probability that a student selected at random scored between 20% and 60%.