Unit -II
Unit -II
Measures of Central tendency
1. Briefly explain measures of Central Tendency ?
Answer - Measures of Central Tendency
A measure of central tendency is a statistical summary that represents the center point of the dataset. It indicates where most values in a distribution fall. It is also called as measure of central location.
The three most common measure of central tendency are Mean, Median, and Mode.
Definition
According to Prof Bowley, “Measures of central tendency (averages) are statistical constants which enable us to comprehend in a single effort the significance of the whole.”
Requisites of a good measure of central tendency
- It should be rigidly defined
- It should be simple to understand and easy to calculate
- It should be based upon all values of given data.
- It should be capable of further mathematical treatment.
- It should have sampling stability.
- It should be not be unduly affected by extreme values.
Mean
- 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.
- Means is calculated by adding up all the values and divided by the number of observation.
Merits of mean
- It is rigidly defined
- It is easy to understand and easy to calculate
- It is based upon all values of the given data
- It is capable of future mathematical treatment
- It is not much affected by sampling fluctuation
Demerits of mean
- It cannot be calculated if any observation are missing
- It cannot be calculated for open end classes
- It is effected by extreme values
- It cannot be located graphically
- It may be number which is not present in the data
Median
- 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
Merits of median
- It is rigidly defined
- It is easy to understand and easy to calculate
- It is not effected 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.
Mode
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”
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.
2. 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
3. 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
4. Calculate the mean number of hours per week spent by each student in texting message.
Time per week | 0 - 5 | 5 - 10 | 10 - 15 | 15 - 20 | 20 - 25 | 25 – 30 |
No. Of students | 8 | 11 | 15 | 12 | 9 | 5 |
Solution:
Time per week (X) | No. Of students (F) | Mid point X | XF |
0 - 5 | 8 | 2.5 | 20 |
5 – 10 | 11 | 7.5 | 82.5 |
10 - 15 | 15 | 12.5 | 187.5 |
15 - 20 | 12 | 17.5 | 210 |
20 - 25 | 9 | 22.5 | 202.5 |
25 – 30 | 5 | 27.5 | 137.5 |
| 60 |
| 840 |
Mean = 840/60 = 14
5. The following table of grouped data represents the weights (in pounds) of all 100 babies born at a local hospital last year.
Weight (pounds) | Number of Babies |
[3−5) | 8 |
[5−7) | 25 |
[7−9) | 45 |
[9−11) | 18 |
[11−13) | 4 |
Solution:
Weight (pounds) | Number of Babies | Mid point X | XF |
[3−5) | 8 | 4 | 32 |
[5−7) | 25 | 6 | 150 |
[7−9) | 45 | 8 | 360 |
[9−11) | 18 | 10 | 180 |
[11−13) | 4 | 12 | 48 |
| 100 |
| 770 |
Mean = 770/100 = 7.7
6. Find the median score of 7 students in science class
Score = 19, 17, 16, 15, 12, 11, 10
Median = (7+1)/2 = 4th value
Median = 15
7. Find the median score of 8 students in science class
Score = 19, 17, 16, 15, 12, 11, 10, 9
Median = (8+1)/2 = 4.5th value
Median = (15+12)/2 = 13.5
8. 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
9. Calculate the median
Marks | No. Of students |
0-4 | 2 |
5-9 | 8 |
10-14 | 14 |
15-19 | 17 |
20-24 | 9 |
Solution:
Marks | No. Of students | CF |
0-4 | 2 | 2 |
5-9 | 8 | 10 |
10-14 | 14 | 24 |
15-19 | 17 | 41 |
20-24 | 9 | 50 |
| 50 |
|
n = 50
n = 50/2= 25
2
The category containing n/2 is 15 -19
Lb = 15
Cfp = 24
f = 17
Ci = 4
Median = 15 + 25-24 *4 = 15.23
17
10. Given the below frequency table calculate median
X | 60 – 70 | 70 – 80 | 80- 90 | 90-100 |
F | 4 | 5 | 6 | 7 |
Solution:
X | F | CF |
60 - 70 | 4 | 4 |
70 - 80 | 5 | 9 |
80 - 90 | 6 | 15 |
90 - 100 | 7 | 22 |
n = 22
n = 22/2= 11
2
The category containing n+1/2 is 80 - 90
Lb = 80
Cfp = 9
f = 6
Ci = 10
Median = 80 + 11-9 *10 = 83.33
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 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.
13. Find the mode
Seconds | Frequency |
51 - 55 | 2 |
56 - 60 | 7 |
61 - 65 | 8 |
66 - 70 | 4 |
The group with the highest frequency is the modal group: - 61-65
D1 = 8-7 = 1
D2 = 8-4 = 4
Mode = L1 + (L2 – L1) d1
d1 +d2
mode = 61 + (65-61) 1 = 61+4 (1/5) = 61.8
1+4
Mode = 61.8
14. 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
15. 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
16. Compute the mode from the following frequency distribution
CI | F |
70-71 | 2 |
68-69 | 2 |
66-67 | 3 |
64-65 | 4 |
62-63 | 6 |
60-61 | 7 |
58-59 | 5 |
Solution:
The group with the highest frequency is the modal group: - 60 - 61
D1 = 7 - 6 = 1
D2 = 7 - 5 = 2
Mode = L1 + (L2 – L1) d1
d1 +d2
mode = 60 + (61-60) 1 = 60+1 (1/3) 60.85
1+2
17. What is Geometric Mean? Explain with examples?
Answer - Geometric mean
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
Example 1 – 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
For grouped data
Geometric Mean, GM = Antilog ∑ f logxi
N
Example 2 – 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
Example 3 – calculate geometric mean
Class | Frequency |
2-4 | 3 |
4-6 | 4 |
6-8 | 2 |
8-10 | 1 |
Solution
Class | Frequency | x | Log x | Flogx |
2-4 | 3 | 3 | 1.0986 | 3.2958 |
4-6 | 4 | 5 | 1.2875 | 6.4378 |
6-8 | 2 | 7 | 0.5559 | 3.8918 |
8-10 | 1 | 9 | 0.2441 | 2.1972 |
| 10 |
|
| 15.8226 |
GM = Antilog ∑ f logxi
N
= antilog 15.8226/10
= antilog 1.5823
GM = 4.866
18. What is Harmonic Mean? Explain with examples.
Answer - Harmonic mean
Harmonic mean is quotient of “number of the given values” and “sum of the reciprocals of the given values
For ungrouped data
Example 1 - Calculate the harmonic mean of the numbers 13.2, 14.2, 14.8, 15.2 and 16.1
Solution
X | 1/X |
13.2 | 0.0758 |
14.2 | 0.0704 |
14.8 | 0.0676 |
15.2 | 0.0658 |
16.1 | 0.0621 |
Total | 0.3147 |
H.M of X = 5/0.3147 = 15.88
Example 2 - Find the harmonic mean of the following data {8, 9, 6, 11, 10, 5} ?
X | 1/X |
8 | 0.125 |
9 | 0.111 |
6 | 0.167 |
11 | 0.091 |
10 | 0.100 |
5 | 0.200 |
Total | 0.794 |
H.M of X = 6/0.794 = 7.560
For grouped data
Example 3 - Calculate the harmonic mean for the below data
Marks | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
F | 2 | 3 | 11 | 20 | 32 | 25 | 7 |
Solution
Marks | X | F | F/X |
30-39 | 34.5 | 2 | 0.0580 |
40-49 | 44.5 | 3 | 0.0674 |
50-59 | 54.4 | 11 | 0.2018 |
60-69 | 64.5 | 20 | 0.3101 |
70-79 | 74.5 | 32 | 0.4295 |
80-89 | 84.5 | 25 | 0.2959 |
90-99 | 94.5 | 7 | 0.0741 |
Total |
| 100 | 1.4368 |
HM = 100/1.4368 = 69.59
Example 4 – find the harmonic mean of the given class
Ages | 4 | 5 | 6 | 7 |
No. Of students | 6 | 4 | 10 | 9 |
Solution
X | F | f/x |
4 | 6 | 1.50 |
5 | 4 | 0.80 |
6 | 10 | 1.67 |
7 | 9 | 1.29 |
| 29.00 | 5.25 |
HM = 29/5.25 = 5.5
Example 5 – calculate harmonic mean
Class | Frequency |
2-4 | 3 |
4-6 | 4 |
6-8 | 2 |
8-10 | 1 |
Solution
Class | Frequency | x | f/x |
2-4 | 3 | 3 | 1 |
4-6 | 4 | 5 | 0.8 |
6-8 | 2 | 7 | 0.28 |
8-10 | 1 | 9 | 0.11 |
| 10 |
| 2.19 |
Harmonic mean = 10/2.19 = 4.55
19. What do you understand by Quartiles, Deciles and Percentile?
Answer - Quartiles
There are three quartiles, i.e. Q1, Q2 and Q3 which divide the total data into four equal parts when it has been orderly arranged. Q1, Q2 and Q3 are termed as first quartile, second quartile and third quartile or lower quartile, middle quartile and upper quartile, respectively. The first quartile, Q1, separates the first one-fourth of the data from the upper three fourths and is equal to the 25th percentile. The second quartile, Q2, divides the data into two equal parts (like median) and is equal to the 50th percentile. The third quartile, Q3, separates the first three-quarters of the data from the last quarter and is equal to 75th percentile.
Calculation of Quartiles:
The calculation of quartiles is done exactly in the same manner as it is in case of the calculation of median.
The different quartiles can be found using the formula given below:
Qi = l1 + 
i= 1,2,3
Where,
L1 = lower limit of ith quartile class
L2 = upper limit of ith quartile class
c = cumulative frequency of the class preceding the ith quartile class
f = frequency of ith quartile class.
Deciles
Deciles are the partition values which divide the arranged data into ten equal parts. There are nine deciles i.e. D1, D2, D3……. D9 and 5th decile is same as median or Q2, because it divides the data in two equal parts.
Calculation of Deciles:
The calculation of deciles is done exactly in the same manner as it is in case of calculation of median.
The different deciles can be found using the formula given below:
Di = l1 + 
i= 1,2,3….9
Where,
l1 = lower limit of ith quartile class
l2 = upper limit of ithquartile class
c = cumulative frequency of the class preceding the ithquartile class
f = frequency of ith quartile class.
Percentile
Percentiles are the values which divide the arranged data into hundred equal parts. There are 99 percentiles i.e. P1, P2, P3, ……. P99.
The 50th percentile divides the series into two equal parts and P50 = D5 = Median.
Similarly, the value of Q1 = P25 and value of Q3 = P75
Calculation of Percentiles:
The different percentiles can be found using the formula given below:
pi = l1 + 
i= 1,2,3…………….99
Where,
l1 = lower limit of ith quartile class
l2 = upper limit of ithquartile class
c = cumulative frequency of the class preceding the ithquartile class
f = frequency of ith quartile class.
20. Calculate Q1, Q2 and Q3 from the following data given below:
Day | Frequency |
1 | 20 |
2 | 35 |
3 | 25 |
4 | 12 |
5 | 10 |
6 | 23 |
7 | 18 |
8 | 14 |
9 | 30 |
10 | 40 |
ANS
Arrange the frequency data in ascending order
Day | Frequency |
1 | 10 |
2 | 12 |
3 | 14 |
4 | 18 |
5 | 20 |
6 | 23 |
7 | 25 |
8 | 30 |
9 | 35 |
10 | 40 |
First quartile (Q1)
Qi= [i * (n + 1) /4] th observation
Q1= [1 * (10 + 1) /4] th observation
Q1 = 2.75 th observation
Thus, 2.75 th observation lies between the 2nd and 3rd value in the ordered group, between frequency 12 and 14
First quartile (Q1) is calculated as
Q1 = 2nd observation +0.75 * (3rd observation - 2nd observation)
Q1 = 12 + 0.75 * (14 – 12) = 13.50
Third quartile (Q3)
Qi= [i * (n + 1) /4] th observation
Q3= [3 * (10 + 1) /4] th observation
Q3 = 8.25 th observation
So, 8.25 th observation lies between the 8th and 9th value in the ordered group, between frequency 30 and 35
Third quartile (Q3) is calculated as
Q3 = 8th observation +0.25 * (9th observation – 8th observation)
Q3 = 30 + 0.25 * (35 – 30) = 31.25
21. Calculate Q1, D7 and P20 from the following data:
3, 13, 11, 11, 5, 4, 2
ANS
Arranging observations in the ascending order we get
2, 3, 4, 5, 11, 11, 13
Here, n = 7
Q1 = (
)th value of the observation
= (
)th Value of the observation
= 2nd Value of the observation
= 3
D3 = (
)th value of the observation
= (
)th value of the observation
= (2.4)th Value of the observation
= 2nd observation + 0.4 (3rd – 2nd)
= 3 + 0.4(4 – 3)
= 3 + 0.4
= 3.4
P20 = (
)th value of the observation
= (
)th value of the observation
= (1.6)th value of the observation
= 1st observation + 0.6 (2nd – 1st )
= 2 + 0.6(3 – 2)
= 2 + 0.6
= 2.6
22. Calculate P20 from the following data:
Class | 2 - 4 | 4 - 6 | 6 - 8 | 8 - 10 |
Frequency | 3 | 4 | 2 | 1 |
ANS
In the case of Frequency Distribution, Percentiles can be calculated by using the formula:
pi = l1 + 
Class interval | F | CF |
2 - 4 | 3 | 3 |
4 - 6 | 4 | 7 |
6 - 8 | 2 | 9 |
8 - 10 | 1 | 10 |
Total | n = 10 |
|
Here n = 10
Class with 
th value of the observation in CF column
= 
th value of the observation in CF column
= 2th value of the observation in CF column and it lies in the class 2 - 4
Therefore, P20 class is 2 – 4
The lower boundary point of 2 – 4 is 2.
Therefore, L = 2
P20 = L + 
= 2 + 
x 2
= 2 + 1.3333
= 3.3333
23. Calculate D7 from the following data:
Class | 2 - 4 | 4 - 6 | 6 - 8 | 8 - 10 |
Frequency | 3 | 4 | 2 | 1 |
ANS
In the case of Frequency Distribution, Deciles can be calculated by using the formula:
Di = l1 + 
Class interval | F | CF |
2 - 4 | 3 | 3 |
4 - 6 | 4 | 7 |
6 - 8 | 2 | 9 |
8 - 10 | 1 | 10 |
Total | n = 10 |
|
Here n = 10
Class with 
th value of the observation in CF column
= 
th value of the observation in CF column
= 7th value of the observation in CF column and it lies in the class 6 – 8
Therefore, D7 class is 6 – 8
The lower boundary point of 6 – 8 is 6.
Therefore, L = 6
D7 = L + 
= 6 + 
x 2
= 6 + 0
= 6
24. Calculate Q1, Q2 and Q3 from the following data given below:
Age in years | 40 -44 | 45 – 49 | 50 – 54 | 55 - 59 | 60 – 64 | 65 - 69 |
Employees | 5 | 8 | 11 | 10 | 9 | 7 |
ANS
In the case of Frequency Distribution, Quartiles can be calculated by using the formula:
Class interval | F | Class boundaries | CF |
40 -44 | 5 | 39.5 – 44.5 | 5 |
45 – 49 | 8 | 44.5 – 49.5 | 13 |
50 – 54 | 11 | 49.5 – 54.5 | 24 |
55 – 59 | 10 | 54.5 – 59.5 | 34 |
60 – 64 | 9 | 59.5 – 64.5 | 43 |
65 – 69 | 7 | 64.5 – 69.5 | 50 |
Total | 50 |
|
|
First quartile (Q1)
Qi= [i * (n ) /4] th observation
Q1 = [1*(50)/4]th observation
Q1 = 12.50th observation
So, 12.50th value is in the interval 44.5 – 49.5
Group of Q1 = 44.5 – 49.5
Qi = (I + (h / f) * ( i * (N/4) – c) ; i = 1,2,3
Q1 = (44.5 + ( 5/ 8)* (1* (50/4) – 5)
Q1 = 49.19
Third quartile (Q3)
Qi= [i * (n) /4] th observation
Q3= [3 * (50) /4] th observation
Q3 = 37.5th observation
So, 37.5th value is in the interval 59.5 – 64.5
Group of Q3 = 59.5 – 64.5
Qi = (I + (h / f) * ( i * (N/4) – c) ; i = 1,2,3
Q3 = (59.5 + ( 5/ 9)* (3* (50/4) – 34)
Q3 = 61.44
