Unit 2
Statistical Averages
Question Bank
- Define Arithmetic Mean. State its merits and demerits.
Solution: The mean is the arithmetic average, also called as arithmetic mean. Mean is very simple to calculate and is most commonly used measure of the center of data. Mean is calculated by adding up all the values and divided by the number of observations.
Computation of sample mean -
If X1, X2, ………………Xn are data values then arithmetic mean is given by
X = =
Computation of the mean for ungrouped data
If X1, X2, ………………Xn are data values and f1, f2,……., fn are its respective frequencies, then arithmetic mean is given by
X =
Merits of Arithmetic Mean
- It is rigidly defined by Arithmetic formulae.
- It is easy to calculate and simple to understand.
- It is based on all observations.
- It is capable of being treated mathematically and hence it is widely used.
Demerits of Arithmetic Mean
- It cannot be determined by inspection or by graphical location.
- It cannot be computed for qualitative data like honesty, intelligence, etc.
- It cannot be computed when class intervals are open.
2. Define Median. State its merits and demerits.
Solution: The points or value that divides the data into two equal parts. Firstly, the data are arranged in ascending or descending order. The median is the middle number depending on the data size. When the data size is odd, the median is the middle value. When the data size is even, median is the average of the middle two values. It is also known as middle score or 50th percentile.
For ungrouped data median is calculated by (n+1) th value
2
Median of grouped data
Formula
MC = median class is a category containing the n/2
Lb = lower boundary of the median class
Cfp = cumulative frequency before the median class if the scores are arranged from lowest to highest value
Fm = frequency of the median class
c.i = size of the class interval
Ex- calculate the median
Merits of median
- It is rigidly defined
- It is easy to understand and easy to calculate
- It is not affected by extreme values
- It is not much affected by sampling fluctuation
- It can be located graphically
Demerits of median
- It is not based upon all values of the given data
- It is difficult to calculate increasing order data size
- It is not capable of further mathematical treatment.
3. What do you mean by Mode? State its merits and demerits.
Solution:The mode is denoted Mo, is the value which occurs most frequently in a set of values. Croxton and Cowden defined it as “the mode of a distribution is the value at the point armed with the item tends to most heavily concentrated. It may be regarded as the most typical of a series of value”.
Mode for Mode for ungrouped data
Example 1- Find the mode of scores of section A
Scores = 25, 24, 24, 20, 17, 18, 10, 18, 9, 7
Solution – Mode is 24, 18 as both have occurred twice.
Mode for grouped data
Mode = L1 + (L2 – L1) d1
d1 +d2
L1= lower limit of the modal class,
L2= upper limit of the modal class‟
d1 =fm-f0 and d2=fm-f1
Where fm= frequency of the modal class,
f0 = frequency of the class preceding to the modal class,
f1= frequency of the class succeeding to the modal class.
Merits of mode
- It is easy to understand & easy to calculate
- It is not affected by extreme values or sampling fluctuations.
- Even if extreme values are not known mode can be calculated.
- It can be located just by inspection in many cases.
- It is always present within the data.
Demerits of mode
- It is not rigidly defined.
- It is not based upon all values of the given data.
- It is not capable of further mathematical treatment.
4. Define Geometric Mean.
Solution: Geometric mean is a type of mean or average, which indicates the central tendency of a set of numbers by using the product of their values.
Definition
The Geometric Mean (G.M) of a series containing n observations is the nth root of the product of the values.
For ungrouped data
Geometric Mean, GM = Antilog ∑logxi
N
5. The marks obtained in 10 class tests are 25, 10, 15, 30, and 35. Calculate Mean
Solution: The mean =
= = = 23
Analysis – The average performance of 5 students is 23. The implication is that students who got below 23 did not perform well. The students who got above 23 performed well in exam.
6. Find the mean:
Xi | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
Freq (Fi) | 2 | 5 | 12 | 17 | 14 | 6 | 3 |
Solution:
Xi | Freq (Fi) | XiFi |
9 | 2 | 18 |
10 | 5 | 50 |
11 | 12 | 132 |
12 | 17 | 204 |
13 | 14 | 182 |
14 | 6 | 84 |
15 | 3 | 45 |
| Fi = 59 | XiFi= 715 |
|
|
|
Then, N = ∑ fi = 59, and ∑fi Xi=715
X = 715/59 = 12.11
7. The following data represent the income distribution of 100 families. Calculate mean income of 100 families?
Income | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |
No. Of families | 8 | 12 | 25 | 22 | 16 | 11 | 6 |
Solution:
Income | No. Of families | Xm (Mid-point) | FXm |
30-40 | 8 | 35 | 280 |
40-50 | 12 | 34 | 408 |
50-60 | 25 | 55 | 1375 |
60-70 | 22 | 65 | 1430 |
70-80 | 16 | 75 | 1200 |
80-90 | 11 | 85 | 935 |
90-100 | 6 | 95 | 570 |
| n = 100 |
| ∑f Xm = 6198 |
X = ∑f Xm/n = 6330/100 = 63.30
Mean = 63.30
8. In a class of 30 students marks obtained by students in science out of 50 is tabulated below. Calculate the mode of the given data.
Marks obtained | No. Of students |
10 -20 | 5 |
20 – 30 | 12 |
30 – 40 | 8 |
40 - 50 | 5 |
Solution:
The group with the highest frequency is the modal group: - 20 -30
D1 = 12 - 5 = 7
D2 = 12 - 8 = 4
Mode = L1 + (L2 – L1) d1
d1 +d2
Mode = 20 + (30-20) 7 = 20+10 (7/11) = 26.36
7+4
Mode = 61.8
9. Based on the group data below, find the mode
Time to travel to work | Frequency |
1 – 10 | 8 |
11 -20 | 14 |
21 – 30 | 12 |
31 – 40 | 9 |
41 - 50 | 7 |
Solution:
The group with the highest frequency is the modal group: - 11 - 20
D1 = 14 - 8 = 6
D2 = 14 - 12 = 2
Mode = L1 + (L2 – L1) d1
d1 +d2
Mode = 11 + (20-11) 6 = 11+9 (6/8) = 17.75
6+2
10. Find the median of the table given below:
Marks obtained | No. Of students |
20 | 6 |
25 | 20 |
28 | 24 |
29 | 28 |
33 | 15 |
38 | 4 |
42 | 2 |
43 | 1 |
Solution:
Marks obtained | No. Of students | Cf |
20 | 6 | 6 |
25 | 20 | 26 (20+6) |
28 | 24 | 50 (26+24) |
29 | 28 | 78 |
33 | 15 | 93 |
38 | 4 | 97 |
42 | 2 | 99 |
43 | 1 | 100 |
Median = (n+1)/2 = 100+1/2 = 50.5
Median = (28+29)/2 = 28.5
11. Calculate the median of grouped data
Class interval | 1-3 | 3-5 | 5-7 | 7-9 | 9-11 | 11-13 |
Frequency | 4 | 12 | 13 | 19 | 7 | 5 |
Solution:
CI | F | CF |
1-3 | 4 | 4 |
3-5 | 12 | 16 |
5-7 | 13 | 29 |
7-9 | 19 | 48 |
9-11 | 7 | 55 |
11-13 | 5 | 60 |
n = 60
n = 60/2= 30
2
The category containing n+1/2 is 7-9
Lb = 7
Cfp = 29
f = 19
Ci = 2
Median = 7 + 30-29 *2 = 7.105
19
12. Find the G.M of the values
X | Log X |
45 | 1.653 |
60 | 1.778 |
48 | 1.681 |
65 | 1.813 |
Total | 6.925 |
GM = Antilog ∑logxi
N
= Antilog 6.925/4
= Antilog 1.73
= 53.82
13. Calculate the geometric mean
X | f |
60 – 80 | 22 |
80 – 100 | 38 |
100 – 120 | 45 |
120 – 140 | 35 |
|
|
Solution:
X | f | Mid X | Log X | f log X |
60 – 80 | 22 | 70 | 1.845 | 40.59 |
80 – 100 | 38 | 90 | 1.954 | 74.25 |
100 – 120 | 45 | 110 | 2.041 | 91.85 |
120 – 140 | 35 | 130 | 2.114 | 73.99 |
Total | 140 |
|
| 280.68 |
GM = Antilog ∑ f logxi
N
= antilog 280.68/140
= antilog 2.00
GM = 100
.