In the previous section we discussed about median. We also saw some examples. In this section we will see a few more examples.
Example 11
The median of the following data is 525. Find the values of x and y, if the total frequency is 100.
Solution:
1. In the given data, two frequencies are missing. They are given as 'x' and 'y'
• However, we will proceed as usual, as if to find the median
• The first step to find the median, is to fill up the cumulative frequency column. It is shown in table 37.44 below:
2. The median is given to us as 525. It falls within the class 500-600
• So 500-600 is the median class
3. Now we can apply the formula:
■ The median is calculated using the following formula:
Where:
Median = 525 (Already given in this problem)
l = lower limit of the median class = 500
n = number of observations = total of the frequency column = (76+x+y) = 100 (given)
cf = cumulative frequency of the class preceding the median class = (36+x)
f = frequency of the median class = 20
h = width of the class interval (assuming all classes are of the same width) = 100
• Substituting the values, we get:
Numerator = (n⁄2 - cf) = [100⁄2 - (36+x)] = [50-36-x] = [14-x]
• Thus median = 525 = 500 + [(14-x)⁄20]×100
⟹ 525 = 500 + 5(14-x)
⟹ 25 = 5(14-x) ⟹ 14-x = 5 ⟹ x = 9
4. We are given that total frequency = 100
• So we can write: 76+x+y = 100 ⟹ x+y = 24
• Substituting x = 9, we get:
y = 24 - 9 = 15
Now we will see some solved examples
Solved example 37.11
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
Solution:
• We will form the complete table from which we can find all the three values: mean, median and mode.
• It is the same type of table that we saw in the case of mean. The only change is the extra column for 'cumulative frequency'. It is shown in magenta colour:
1. First we will find the mean
We will use the step deviation method:
• u is given by the formula:
• The numerator is the value at the bottom end of the eighth column. It is 43
• The denominator is the value at the bottom end of the second column. It is 80
• So we get u = 7⁄68
• Thus x = a + hu = [135 + (20 × 7⁄68)] = 137.06
2. Now we will find the mode
(i) The modal class is the class interval having the highest frequency
• So in this problem, the class interval '125- 145' is the modal class. It has the highest frequency of '20'
(ii) Now we can calculate the mode using the formula:
l = lower limit of the modal class = 125
h = width of the class interval (assuming all classes are of the same width) = 20
f1 = frequency of the modal class = 20
f0 = frequency of the class preceding the modal class = 13
f2 = frequency of the class succeeding the modal class = 14
Substituting all the values, we get:
mode = 125 + (20-13⁄2×20-13-14)×20 = 125 + (7⁄40-27)×20 = 135.77
3. Now we will calculate the median:
• In this problem, the total number of observations 'n' = 68. This is an even number.
So n⁄2 = 68⁄2 = 34
• The cumulative frequency closest to and just greater than 34 is 42
• The class corresponding to the cumulative frequency 42 is 125-145
• So 125-145 is the median class. It's frequency = 20
■ The median is calculated using the following formula:
Where:
l = lower limit of the median class = 125
n = number of observations = 68
cf = cumulative frequency of the class preceding the median class = 22
f = frequency of the median class = 20
h = width of the class interval (assuming all classes are of the same width) = 20
• Substituting the values, we get:
• Numerator = (n⁄2 - cf) = (68⁄2 - 22) = (34-22) = 12
• Thus median = 125 + (12⁄20)×20 = 125 + 12 = 137
■ We can write the conclusion:
1. Mean is obtained as 137.06 units:
(i) The average consumption of electricity in that locality in a month = 137.06
(ii) There are 68 families. So the total consumption will be about (137.06 × 68) = 9320.08 units
(iii) Even if there are more than 68 families, this mean can be used.
(iv) For example, if there are 450 families, the total consumption in a month in that locality will be (137.06 × 450) = 61677 units
2. Mode is obtained as 135.77:
(i) Among the 68 families, 'number of families consuming 135.77 units per month' is more than others
(ii) If we list the monthly consumption of each of the 68 families individually, the 'value 135.77 units' will occur more no. of times than others
3. Median is obtained as 137 units
(i) If we list the monthly consumption of all the 68 families in the ascending order, 137 will occupy nearly the 34th position.
(ii) The values to the left of 137 will be less and the values to the right will be more
Solved example 37.12
If the median of the distribution given below is 28.5, find the values of x and y
Solution:
1. In the given data, two frequencies are missing. They are given as 'x' and 'y'
• However, we will proceed as usual, as if to find the median
• The first step to find the median, is to fill up the cumulative frequency column. It is shown in table 37.48 below:
2. The median is given to us as 28.5. It falls within the class 20-30
• So 20-30 is the median class
3. Now we can apply the formula:
■ The median is calculated using the following formula:
Where:
Median = 28.5 (Already given in this problem)
l = lower limit of the median class = 20
n = number of observations = total of the frequency column = (45+x+y) = 60 (given)
cf = cumulative frequency of the class preceding the median class = (5+x)
f = frequency of the median class = 20
h = width of the class interval (assuming all classes are of the same width) = 10
• Substituting the values, we get:
Numerator = (n⁄2 - cf) = [60⁄2 - (5+x)] = [30-5-x] = [25-x]
• Thus median = 28.5 = 20 + [(25-x)⁄20]×10
⟹ 28.5 = 20 + 1⁄2 × (25-x)
⟹ 8.5 = 1⁄2 × (25-x) ⟹ 17 = (25-x) ⟹ x = 8
4. We are given that total frequency = 60
• So we can write: 45+x+y = 60 ⟹ x+y = 15
• Substituting x = 8, we get:
y = 15 - 8 = 7
Solved example 37.13
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years on wards but less than 60 year
Solution:
1. The given data is in the form of a cumulative frequency table.
We have to convert it into a 'grouped frequency distribution table'. Let us write the steps:
• No. of policy holders below age 20 is 2
• No. of policy holders below age 25 is 6
• This 6 will include the 'no. of policy holders of ages below 20' also
2. From the above three information, we can find the 'no. of policy holders whose ages satisfy the following two conditions':
♦ No. of ages greater than or equal to 20
♦ No. of ages less than 25
• Obviously, the no. of ages satisfying the two conditions is: (6-2) = 4
• So 4 policy holders satisfy the condition: 20 ≤ age < 25
• Those 4 policy holders will obviously fall in the class: 20-25
• In other words, the frequency of the class 20-25 is 4
3. So we separated out 4 policy holders from 6
What about the remaining 2?
• They satisfy the condition: 'age < 20'
• Those 2 policy holders will obviously fall in the class: k-20
♦ Where 'k' is any number less than 20. We will give it a value 15
• So the frequency of the class 15-20 is 2
4. So we found two classes:
♦ class 15-20 with frequency 2
♦ class 20-25 with frequency 4
Now we consider the next cumulative frequency, which is 24:
• No. of policy holders below age 25 is 6
• No. of policy holders below age 30 is 24
• This 24 will include the 'no. of policy holders of ages below 25' also
• From the above three information, we can find the 'no. of policy holders whose ages satisfy the following two conditions':
♦ No. of ages greater than or equal to 25
♦ No. of ages less than 30
• Obviously, the no. of ages satisfying the two conditions is: (24-6) = 18
• So 18 policy holders satisfy the condition: 25 ≤ age < 30
• Those 18 policy holders will obviously fall in the class: 25-30
• In other words, the frequency of the class 25-30 is 18
5. Thus so far, we found three classes:
♦ class 15-20 with frequency 2
♦ class 20-25 with frequency 4
♦ class 25-30 with frequency 18
Now we consider the next cumulative frequency, which is 45:
• No. of policy holders below age 30 is 24
• No. of policy holders below age 35 is 45
• This 45 will include the 'no. of policy holders of ages below 30' also
• From the above three information, we can find the 'no. of policy holders whose ages satisfy the following two conditions':
♦ No. of ages greater than or equal to 30
♦ No. of ages less than 35
• Obviously, the no. of ages satisfying the two conditions is: (45-24) = 21
• So 21 policy holders satisfy the condition: 30 ≤ age < 35
• Those 21 policy holders will obviously fall in the class: 30-35
• In other words, the frequency of the class 30-35 is 21
• Thus so far, we found four classes:
♦ class 15-20 with frequency 2
♦ class 20-25 with frequency 4
♦ class 25-30 with frequency 18
♦ class 30-35 with frequency 21
■ We can go on like this until we reach the end of the table. However, once we understand the pattern, the frequencies of each class can be easily calculated. The resulting table is shown below:
• Note how easy it is to fill up the 'frequency column':
To fill up any one row in the 'frequency column', do these steps:
(i) Take the value coming in the same row in the 'cumulative frequency column'
(ii) Take the value coming in the row just above in the 'cumulative frequency column'
(iii) Subtract (ii) from (i)
6. Now we can begin the calculations for the median:
• In this problem, the total number of observations 'n' = 100. This is an even number.
So n⁄2 = 100⁄2 = 50
• The cumulative frequency closest to 50 and greater than 50 is 78
• The class corresponding to the cumulative frequency 78 is 35-40
• So 35-40 is the median class. It's frequency = 33
■ The median is calculated using the following formula:
Where:
l = lower limit of the median class = 35
n = number of observations = 100
cf = cumulative frequency of the class preceding the median class = 45
f = frequency of the median class = 33
h = width of the class interval (assuming all classes are of the same width) = 5
• Substituting the values, we get:
• Numerator = (n⁄2 - cf) = (100⁄2 - 45) = (50-45) = 5
• Thus median = 35 + (5⁄33)×5 = 35 + 0.76 = 35.76
7. We can write the conclusion:
(i) If we list the ages of all the 100 policy holders in the ascending order, 35.76 will occupy nearly the 50th position.
(ii) The ages to the left of 35.67 will be less and the ages to the right will be more
Solved example 37.14
The lengths of 40 leaves of a plant are measured correct to the nearest mm, and the data obtained is represented in the following table:
Find the median length of the leaves
Solution:
1. The class intervals in the given table are not continuous. We can apply the formula for median only if the class intervals are continuous.
We know how to make them continuous (Details here).
So the new class intervals are given in table 37.52 below. The cumulative frequencies are also shown along with it.
2. Now we can begin the calculations for the median:
• In this problem, the total number of observations 'n' = 40. This is an even number.
So n⁄2 = 40⁄2 = 20
• The cumulative frequency closest to 20 and greater than 20 is 29
• The class corresponding to the cumulative frequency 29 is 144.5-153.5
• So 144.5-153.5 is the median class. It's frequency = 12
■ The median is calculated using the following formula:
Where:
l = lower limit of the median class = 144.5
n = number of observations = 40
cf = cumulative frequency of the class preceding the median class = 17
f = frequency of the median class = 12
h = width of the class interval (assuming all classes are of the same width) = 9
• Substituting the values, we get:
• Numerator = (n⁄2 - cf) = (40⁄2 - 17) = (20-17) = 3
• Thus median = 144.5 + (3⁄12)×9 = 144.5 + 2.25 = 146.75
3. We can write the conclusion:
(i) If we arrange all the 40 leaves in the ascending order of their lengths, the leaf with 146.75 mm length will occupy nearly the 20th position.
■ We have discussed about the three measures of central tendency: Mean, mode and median. Each of them have their merits and demerits. So they are used according to the 'end result to be achieved'.
• Let us see an example:
• Mean of five lengths 32, 35, 30, 39 and 38 is: (174⁄5) = 34.8 cm
• If a sixth length 34 cm is also added, the mean of 6 heights is: (208⁄6) = 34.67
• If the sixth height added is 2 instead of 34, the mean will be (176⁄6) = 29.33
• When the sixth length added is 34, the mean changed from 34.8 to 34.67. This is only a mild change
• When the sixth length added is 2 instead of 34, the mean changed from 34.8 to 29.33. This is a significant change
• So we see that extreme values in a group will in influence the mean.
• In other words, if extreme values are present, the mean will not give the true nature of the data
• At the same time, it may not be permissible to discard the extreme values.
• So the decision about 'using mean or median' depends on the situation
• If individual values need not be presented in the result, we can use median. It gives an accurate value at the middle. Again it depends on the situation
■ Mode is used when we want the most repeating value. Examples are:
Mostly watched TV shows, Mostly preferred shoes, Mostly preferred colour for car etc.,
■ There is a empirical relationship between the three measures of central tendency :
3 Median = Mode + 2 Mean
Let us check:
1. Consider the solved example 37.11 above. In that example, we obtained all the results:
• mean = 137.06
• mode = 135.77
• median = 137
2. Left side = 3 times median = 3 × 137 = 411
3. Right side = (mode + 2 times mean) = (135.77 + 2 × 137.06) = (135.77 + 274.12) = 409.89
• The two sides are approximately equal
In the next section, we will see ogive curves.
Example 11
The median of the following data is 525. Find the values of x and y, if the total frequency is 100.
 |
| Table.37.43 |
1. In the given data, two frequencies are missing. They are given as 'x' and 'y'
• However, we will proceed as usual, as if to find the median
• The first step to find the median, is to fill up the cumulative frequency column. It is shown in table 37.44 below:
 |
| Table.37.44 |
• So 500-600 is the median class
3. Now we can apply the formula:
■ The median is calculated using the following formula:
Median = 525 (Already given in this problem)
l = lower limit of the median class = 500
n = number of observations = total of the frequency column = (76+x+y) = 100 (given)
cf = cumulative frequency of the class preceding the median class = (36+x)
f = frequency of the median class = 20
h = width of the class interval (assuming all classes are of the same width) = 100
• Substituting the values, we get:
Numerator = (n⁄2 - cf) = [100⁄2 - (36+x)] = [50-36-x] = [14-x]
• Thus median = 525 = 500 + [(14-x)⁄20]×100
⟹ 525 = 500 + 5(14-x)
⟹ 25 = 5(14-x) ⟹ 14-x = 5 ⟹ x = 9
4. We are given that total frequency = 100
• So we can write: 76+x+y = 100 ⟹ x+y = 24
• Substituting x = 9, we get:
y = 24 - 9 = 15
Now we will see some solved examples
Solved example 37.11
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
 |
| Table.37.45 |
• We will form the complete table from which we can find all the three values: mean, median and mode.
• It is the same type of table that we saw in the case of mean. The only change is the extra column for 'cumulative frequency'. It is shown in magenta colour:
 |
| Table.37.46 |
We will use the step deviation method:
• u is given by the formula:
• The denominator is the value at the bottom end of the second column. It is 80
• So we get u = 7⁄68
• Thus x = a + hu = [135 + (20 × 7⁄68)] = 137.06
2. Now we will find the mode
(i) The modal class is the class interval having the highest frequency
• So in this problem, the class interval '125- 145' is the modal class. It has the highest frequency of '20'
(ii) Now we can calculate the mode using the formula:
h = width of the class interval (assuming all classes are of the same width) = 20
f1 = frequency of the modal class = 20
f0 = frequency of the class preceding the modal class = 13
f2 = frequency of the class succeeding the modal class = 14
Substituting all the values, we get:
mode = 125 + (20-13⁄2×20-13-14)×20 = 125 + (7⁄40-27)×20 = 135.77
3. Now we will calculate the median:
• In this problem, the total number of observations 'n' = 68. This is an even number.
So n⁄2 = 68⁄2 = 34
• The cumulative frequency closest to and just greater than 34 is 42
• The class corresponding to the cumulative frequency 42 is 125-145
• So 125-145 is the median class. It's frequency = 20
■ The median is calculated using the following formula:
Where:
l = lower limit of the median class = 125
n = number of observations = 68
cf = cumulative frequency of the class preceding the median class = 22
f = frequency of the median class = 20
h = width of the class interval (assuming all classes are of the same width) = 20
• Substituting the values, we get:
• Numerator = (n⁄2 - cf) = (68⁄2 - 22) = (34-22) = 12
• Thus median = 125 + (12⁄20)×20 = 125 + 12 = 137
■ We can write the conclusion:
1. Mean is obtained as 137.06 units:
(i) The average consumption of electricity in that locality in a month = 137.06
(ii) There are 68 families. So the total consumption will be about (137.06 × 68) = 9320.08 units
(iii) Even if there are more than 68 families, this mean can be used.
(iv) For example, if there are 450 families, the total consumption in a month in that locality will be (137.06 × 450) = 61677 units
2. Mode is obtained as 135.77:
(i) Among the 68 families, 'number of families consuming 135.77 units per month' is more than others
(ii) If we list the monthly consumption of each of the 68 families individually, the 'value 135.77 units' will occur more no. of times than others
3. Median is obtained as 137 units
(i) If we list the monthly consumption of all the 68 families in the ascending order, 137 will occupy nearly the 34th position.
(ii) The values to the left of 137 will be less and the values to the right will be more
Solved example 37.12
If the median of the distribution given below is 28.5, find the values of x and y
 |
| Table.37.47 |
1. In the given data, two frequencies are missing. They are given as 'x' and 'y'
• However, we will proceed as usual, as if to find the median
• The first step to find the median, is to fill up the cumulative frequency column. It is shown in table 37.48 below:
 |
| Table.37.48 |
• So 20-30 is the median class
3. Now we can apply the formula:
■ The median is calculated using the following formula:
Median = 28.5 (Already given in this problem)
l = lower limit of the median class = 20
n = number of observations = total of the frequency column = (45+x+y) = 60 (given)
cf = cumulative frequency of the class preceding the median class = (5+x)
f = frequency of the median class = 20
h = width of the class interval (assuming all classes are of the same width) = 10
• Substituting the values, we get:
Numerator = (n⁄2 - cf) = [60⁄2 - (5+x)] = [30-5-x] = [25-x]
• Thus median = 28.5 = 20 + [(25-x)⁄20]×10
⟹ 28.5 = 20 + 1⁄2 × (25-x)
⟹ 8.5 = 1⁄2 × (25-x) ⟹ 17 = (25-x) ⟹ x = 8
4. We are given that total frequency = 60
• So we can write: 45+x+y = 60 ⟹ x+y = 15
• Substituting x = 8, we get:
y = 15 - 8 = 7
Solved example 37.13
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years on wards but less than 60 year
 |
| Table.37.49 |
1. The given data is in the form of a cumulative frequency table.
We have to convert it into a 'grouped frequency distribution table'. Let us write the steps:
• No. of policy holders below age 20 is 2
• No. of policy holders below age 25 is 6
• This 6 will include the 'no. of policy holders of ages below 20' also
2. From the above three information, we can find the 'no. of policy holders whose ages satisfy the following two conditions':
♦ No. of ages greater than or equal to 20
♦ No. of ages less than 25
• Obviously, the no. of ages satisfying the two conditions is: (6-2) = 4
• So 4 policy holders satisfy the condition: 20 ≤ age < 25
• Those 4 policy holders will obviously fall in the class: 20-25
• In other words, the frequency of the class 20-25 is 4
3. So we separated out 4 policy holders from 6
What about the remaining 2?
• They satisfy the condition: 'age < 20'
• Those 2 policy holders will obviously fall in the class: k-20
♦ Where 'k' is any number less than 20. We will give it a value 15
• So the frequency of the class 15-20 is 2
4. So we found two classes:
♦ class 15-20 with frequency 2
♦ class 20-25 with frequency 4
Now we consider the next cumulative frequency, which is 24:
• No. of policy holders below age 25 is 6
• No. of policy holders below age 30 is 24
• This 24 will include the 'no. of policy holders of ages below 25' also
• From the above three information, we can find the 'no. of policy holders whose ages satisfy the following two conditions':
♦ No. of ages greater than or equal to 25
♦ No. of ages less than 30
• Obviously, the no. of ages satisfying the two conditions is: (24-6) = 18
• So 18 policy holders satisfy the condition: 25 ≤ age < 30
• Those 18 policy holders will obviously fall in the class: 25-30
• In other words, the frequency of the class 25-30 is 18
5. Thus so far, we found three classes:
♦ class 15-20 with frequency 2
♦ class 20-25 with frequency 4
♦ class 25-30 with frequency 18
Now we consider the next cumulative frequency, which is 45:
• No. of policy holders below age 30 is 24
• No. of policy holders below age 35 is 45
• This 45 will include the 'no. of policy holders of ages below 30' also
• From the above three information, we can find the 'no. of policy holders whose ages satisfy the following two conditions':
♦ No. of ages greater than or equal to 30
♦ No. of ages less than 35
• Obviously, the no. of ages satisfying the two conditions is: (45-24) = 21
• So 21 policy holders satisfy the condition: 30 ≤ age < 35
• Those 21 policy holders will obviously fall in the class: 30-35
• In other words, the frequency of the class 30-35 is 21
• Thus so far, we found four classes:
♦ class 15-20 with frequency 2
♦ class 20-25 with frequency 4
♦ class 25-30 with frequency 18
♦ class 30-35 with frequency 21
■ We can go on like this until we reach the end of the table. However, once we understand the pattern, the frequencies of each class can be easily calculated. The resulting table is shown below:
 |
| Table.37.50 |
To fill up any one row in the 'frequency column', do these steps:
(i) Take the value coming in the same row in the 'cumulative frequency column'
(ii) Take the value coming in the row just above in the 'cumulative frequency column'
(iii) Subtract (ii) from (i)
6. Now we can begin the calculations for the median:
• In this problem, the total number of observations 'n' = 100. This is an even number.
So n⁄2 = 100⁄2 = 50
• The cumulative frequency closest to 50 and greater than 50 is 78
• The class corresponding to the cumulative frequency 78 is 35-40
• So 35-40 is the median class. It's frequency = 33
■ The median is calculated using the following formula:
Where:
l = lower limit of the median class = 35
n = number of observations = 100
cf = cumulative frequency of the class preceding the median class = 45
f = frequency of the median class = 33
h = width of the class interval (assuming all classes are of the same width) = 5
• Substituting the values, we get:
• Numerator = (n⁄2 - cf) = (100⁄2 - 45) = (50-45) = 5
• Thus median = 35 + (5⁄33)×5 = 35 + 0.76 = 35.76
7. We can write the conclusion:
(i) If we list the ages of all the 100 policy holders in the ascending order, 35.76 will occupy nearly the 50th position.
(ii) The ages to the left of 35.67 will be less and the ages to the right will be more
Solved example 37.14
The lengths of 40 leaves of a plant are measured correct to the nearest mm, and the data obtained is represented in the following table:
 |
| Table.37.51 |
Solution:
1. The class intervals in the given table are not continuous. We can apply the formula for median only if the class intervals are continuous.
We know how to make them continuous (Details here).
So the new class intervals are given in table 37.52 below. The cumulative frequencies are also shown along with it.
 |
| Table.37.52 |
• In this problem, the total number of observations 'n' = 40. This is an even number.
So n⁄2 = 40⁄2 = 20
• The cumulative frequency closest to 20 and greater than 20 is 29
• The class corresponding to the cumulative frequency 29 is 144.5-153.5
• So 144.5-153.5 is the median class. It's frequency = 12
■ The median is calculated using the following formula:
Where:
l = lower limit of the median class = 144.5
n = number of observations = 40
cf = cumulative frequency of the class preceding the median class = 17
f = frequency of the median class = 12
h = width of the class interval (assuming all classes are of the same width) = 9
• Substituting the values, we get:
• Numerator = (n⁄2 - cf) = (40⁄2 - 17) = (20-17) = 3
• Thus median = 144.5 + (3⁄12)×9 = 144.5 + 2.25 = 146.75
3. We can write the conclusion:
(i) If we arrange all the 40 leaves in the ascending order of their lengths, the leaf with 146.75 mm length will occupy nearly the 20th position.
■ We have discussed about the three measures of central tendency: Mean, mode and median. Each of them have their merits and demerits. So they are used according to the 'end result to be achieved'.
• Let us see an example:
• Mean of five lengths 32, 35, 30, 39 and 38 is: (174⁄5) = 34.8 cm
• If a sixth length 34 cm is also added, the mean of 6 heights is: (208⁄6) = 34.67
• If the sixth height added is 2 instead of 34, the mean will be (176⁄6) = 29.33
• When the sixth length added is 34, the mean changed from 34.8 to 34.67. This is only a mild change
• When the sixth length added is 2 instead of 34, the mean changed from 34.8 to 29.33. This is a significant change
• So we see that extreme values in a group will in influence the mean.
• In other words, if extreme values are present, the mean will not give the true nature of the data
• At the same time, it may not be permissible to discard the extreme values.
• So the decision about 'using mean or median' depends on the situation
• If individual values need not be presented in the result, we can use median. It gives an accurate value at the middle. Again it depends on the situation
■ Mode is used when we want the most repeating value. Examples are:
Mostly watched TV shows, Mostly preferred shoes, Mostly preferred colour for car etc.,
■ There is a empirical relationship between the three measures of central tendency :
3 Median = Mode + 2 Mean
Let us check:
1. Consider the solved example 37.11 above. In that example, we obtained all the results:
• mean = 137.06
• mode = 135.77
• median = 137
2. Left side = 3 times median = 3 × 137 = 411
3. Right side = (mode + 2 times mean) = (135.77 + 2 × 137.06) = (135.77 + 274.12) = 409.89
• The two sides are approximately equal
No comments:
Post a Comment