Question-1
Calculate Range and Q.D. of the following observations:20, 25, 29, 30, 35, 39, 41, 48, 51, 60 and 70
Solution:
Range is clearly 70 â€“ 20 = 50
For Q.D., we need to calculate values of Q3 and Q1.
Q1 is the size of value. Here, n being 11, Q1 is the size of 3^{rd} value.
As the values are already arranged in ascending order, it can be seen that Q1, the 3rd value is 29. [What will you do if these values are not in an order?]
Similarly, Q3 is size of th value; i.e. 9th value which is 51. Hence Q_{3} = 51
Q.D = = = 11
Do you notice that Q.D. is the average difference of the Quartiles from the median.
Question-2
Problem 2
For the following distribution of marks scored by a class of 40 students, calculate the Range and Q.D.
Class intervals |
No.of students |
0-10 |
5 |
10-20 |
8 |
20-40 |
16 |
40-60 |
7 |
60-90 |
4 |
40 |
Solution:
Range is just the difference between the upper limit of the highest class and the lower limit of the lowest class. So Range is 90 â€“ 0 = 90. For Q.D., first calculate cumulative frequencies as follows:
C.I |
Frequency |
Cumulative Frequency |
0-10 |
5 |
05 |
10-20 |
8 |
13 |
20-40 |
16 |
29 |
40-60 |
7 |
36 |
60-90 |
4 |
40 |
n |
40 |
Q1 is the size of th value in a continuous series. Thus it is the size of the 10th value. The class containing the 10th value is 10â€“20. Hence Q1 lies in class 10â€“20. Now, to calculate the exact value of Q1, the following formula is used:
Where L = 10 (lower limit of the relevant Quartile class) c.f. = 5 (Value of c.f. for the class preceding the Quartile class) i = 10 (interval of the Quartile class), and f = 8 (frequency of the Quartile class) Thus,
= 16.25
Similarly,
Q3 is the size of th value; i.e., 30th value, which lies in class 40â€“60. Now using the formula for Q3, its value can be calculated as follows:
.
Question-3
Solution:
The A M = = 6
X |
{D} |
2 |
4 |
4 |
2 |
7 |
1 |
8 |
2 |
9 |
3 |
12 |
Question-4
Problem 4
X |
{D} |
2 |
5 |
4 |
3 |
7 |
0 |
8 |
1 |
9 |
2 |
11 |
Solution:
Where is the sum of absolute deviations taken from the assumed mean. x is the actual mean. A x is the assumed mean used to calculate deviations.
is the number of values below the actual mean including the actual mean.
is the number of values above the actual mean. Substituting the values in the above formula:
= 2.4
Question-5
Solution:
Range (R) is the difference between the largest (L) and the smallest value (S) in a distribution. Thus,
R = L â€“ S
Question-6
Solution:
The presence of even one extremely high or low value in a distribution can reduce the utility of range as a measure of dispersion. Thus, you need a measure which is not unduly affected by the outliers.
In such a situation, if the entire data is divided into four equal parts, each containing 25% of the values, we get the values of Quartiles and Median. The upper and lower quartiles (Q3 and Q1, respectively) are used to calculate Inter Quartile Range which is Q3 â€“ Q1. Inter-Quartile Range is based upon middle 50% of the values in a distribution and is, therefore, not affected by extreme values. Half of the Inter-Quartile Range is called Quartile Deviation. Thus:
Q .D. =
Q.D. is therefore also called Semi- Inter Quartile Range.
Question-7
Solution:
Recall that dispersion was defined as the extent to which values differ from their average. Range and Quartile Deviation do not attempt to calculate, how far the values are, from their average. Yet, by calculating the spread of values, they do give a good idea about the dispersion. Two measures which are based upon deviation of the values from their average are Mean Deviation and Standard Deviation. Since the average is a central value, some deviations are positive and some are negative. If these are added as they are, the sum will not reveal anything. In fact, the sum of deviations from Arithmetic Mean is always zero.
Question-8
Solution:
Steps:
(i) The A.M. of the values is calculated
(ii) Difference between each value and the A.M. is calculated. All differences are considered positive. These are denoted as |d|
(iii) The A.M. of these differences (called deviations) is the Mean Deviation. i.e. MD =
Question-9
Solution:
Standard Deviation is the positive square root of the mean of squared deviations from mean. So if there are five values x_{1}, x_{2}, x_{3}, x_{4} and x_{5}, first their mean is calculated. Then deviations of the values from mean are calculated. These deviations are then squared. The mean of these squared deviations is the variance. Positive square root of the variance is the standard deviation. (Note that Standard Deviation is calculated on the basis of the mean only).
Question-10
Solution:
Four alternative methods are available for the calculation of standard deviation of individual values. All these methods result in the same value of standard deviation. These are:
(i) Actual Mean Method
(ii) Assumed Mean Method
(iii) Direct Method
(iv) Step-Deviation Method
Following formula to be used for calculating SD
Question-11
Solution:
1. Calculate class mid-points (Col. 3) and deviations from an arbitrarily chosen value, just like in the assumed mean method. In this example, deviations have been taken from the value 40. (Col. 4)
2. Divide the deviations by a common factor denoted as â€˜Câ€™. C = 5 in the above example. The values so obtained are â€˜d'â€™ values (Col. 5).
3. Multiply â€˜d'â€™ values with corresponding â€˜f'â€™ values (Col. 2) to obtain â€˜fd'â€™ values (Col. 6).
4. Multiply â€˜fd'â€™ values with â€˜d'â€™ values to get â€˜fd'2â€™ values (Col. 7)
5. Sum up values in Col. 6 and Col.
7. to get S fd' and S fd'2 values.
Question-12
Solution:
The measures of dispersion discussed so far give a numerical value of dispersion. A graphical measure called Lorenz Curve is available for estimating dispersion. You may have heard of statements like â€˜top 10% of the people of a country earn 50% of the national income while top 20% account for 80%â€™. An idea about income disparities is given by such figures. Lorenz Curve uses the information expressed in a cumulative manner to indicate the degree of variability. It is especially useful in comparing the variability of two or more distributions.
Question-13
Solution:
1. Calculate class mid-points and find cumulative totals as in Col. 3 in the example 16, given above.
2. Calculate cumulative frequencies as in Col. 6.
3. Express the grand totals of Col. 3 and 6 as 100, and convert the cumulative totals in these columns into percentages, as in Col. 4 and 7.
4. Now, on the graph paper, take the cumulative percentages of the variable (incomes) on Y axis and cumulative percentages of frequencies (number of employees) on X-axis, as in figure 6.1. Thus each axis will have values from â€˜0â€™ to â€˜100â€™.
5. Draw a line joining Co-ordinate (0, 0) with (100,100). This is called the line of equal distribution shown as line â€˜OCâ€™ in figure 6.1.
6. Plot the cumulative percentages of the variable with corresponding cumulative percentages of frequency. Join these points to get the curve OAC.