Statistics - Mean, Median, Mode of Grouped Data

Mean of Grouped Data
Mode of Grouped Data
Median of Grouped Data
Exercise Solutions

Mean of Grouped Data

Direct Method

The mean (or average) of observations is the sum of the values of all the observations divided by the total number of observations.

\(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\)

Where:

  • \(x_i\) = class mark (mid-point of class interval)
  • \(f_i\) = frequency of the class
Assumed Mean Method

This method is used when the numerical values of \(x_i\) and \(f_i\) are large.

\(\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}\)

Where:

  • \(a\) = assumed mean
  • \(d_i = x_i - a\)
Step Deviation Method

This method is used when all the \(d_i\)'s have a common factor.

\(\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h\)

Where:

  • \(a\) = assumed mean
  • \(h\) = class size
  • \(u_i = \frac{x_i - a}{h}\)
Example 1: Direct Method

The marks obtained by 30 students of Class X in a Mathematics paper:

Marks obtained (xi) 10 20 36 40 50 56 60 70 72 80 88 92 95
Number of students (fi) 1 1 3 4 3 2 4 4 1 1 2 3 1

Solution:

We calculate \(f_i x_i\) for each class:

Marks obtained (xi) Number of students (fi) fixi
10110
20120
363108
404160
503150
562112
604240
704280
72172
80180
882176
923276
95195
Total \(\sum f_i = 30\) \(\sum f_i x_i = 1779\)

Now, \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1779}{30} = 59.3\)

Therefore, the mean marks obtained is 59.3.

Example 2: All Three Methods

Percentage distribution of female teachers in primary schools:

Percentage of female teachers 15-25 25-35 35-45 45-55 55-65 65-75 75-85
Number of States/U.T. 6 11 7 4 4 2 1

Solution:

First, we find the class marks (\(x_i\)) and then calculate using all three methods:

Class Interval fi xi di = xi - 50 ui = (xi - 50)/10 fixi fidi fiui
15-25620-30-3120-180-18
25-351130-20-2330-220-22
35-45740-10-1280-70-7
45-554500020000
55-65460101240404
65-75270202140404
75-8518030380303
Total 35 1390 -360 -36

Direct Method: \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1390}{35} = 39.71\)

Assumed Mean Method: \(\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 50 + \frac{-360}{35} = 39.71\)

Step Deviation Method: \(\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h = 50 + \left( \frac{-36}{35} \right) \times 10 = 39.71\)

Therefore, the mean percentage of female teachers is 39.71%.

Mode of Grouped Data

Formula for Mode

For grouped data, the mode is calculated using the formula:

\(\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\)

Where:

  • \(l\) = lower limit of the modal class
  • \(h\) = size of the class interval
  • \(f_1\) = frequency of the modal class
  • \(f_0\) = frequency of the class preceding the modal class
  • \(f_2\) = frequency of the class succeeding the modal class
Example 5: Finding Mode

Family size distribution in 20 households:

Family size 1-3 3-5 5-7 7-9 9-11
Number of families 7 8 2 2 1

Solution:

The maximum class frequency is 8, and the class corresponding to this frequency is 3-5. So, the modal class is 3-5.

Now:

  • \(l\) = 3
  • \(h\) = 2
  • \(f_1\) = 8
  • \(f_0\) = 7
  • \(f_2\) = 2

Using the formula:

\(\text{Mode} = 3 + \left( \frac{8 - 7}{2 \times 8 - 7 - 2} \right) \times 2 = 3 + \frac{2}{7} = 3.286\)

Therefore, the mode of the data is 3.286.

Example 6: Comparing Mode and Mean

Marks distribution of 30 students (from Example 1):

Class interval 10-25 25-40 40-55 55-70 70-85 85-100
Number of students 2 3 7 6 6 6

Solution:

The maximum number of students (7) have got marks in the interval 40-55, so the modal class is 40-55.

Now:

  • \(l\) = 40
  • \(h\) = 15
  • \(f_1\) = 7
  • \(f_0\) = 3
  • \(f_2\) = 6

Using the formula:

\(\text{Mode} = 40 + \left( \frac{7 - 3}{14 - 6 - 3} \right) \times 15 = 52\)

From Example 1, the mean marks is 62.

So, the maximum number of students obtained 52 marks, while on average a student obtained 62 marks.

Median of Grouped Data

Formula for Median

For grouped data, the median is calculated using the formula:

\(\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h\)

Where:

  • \(l\) = lower limit of median class
  • \(n\) = total number of observations
  • \(cf\) = cumulative frequency of the class preceding the median class
  • \(f\) = frequency of median class
  • \(h\) = class size
Example: Finding Median

Marks distribution of 53 students:

Marks Number of students Cumulative frequency
0-1055
10-2038
20-30412
30-40315
40-50318
50-60422
60-70729
70-80938
80-90745
90-100853

Solution:

Here \(n = 53\), so \(\frac{n}{2} = 26.5\)

The class whose cumulative frequency is greater than and nearest to 26.5 is 60-70 (cumulative frequency = 29). So, the median class is 60-70.

Now:

  • \(l\) = 60
  • \(n\) = 53
  • \(cf\) = 22
  • \(f\) = 7
  • \(h\) = 10

Using the formula:

\(\text{Median} = 60 + \left( \frac{26.5 - 22}{7} \right) \times 10 = 60 + \frac{45}{7} = 66.4\)

So, about half the students have scored marks less than 66.4, and the other half have scored marks more than 66.4.

Example 7: Median from Cumulative Frequency

Heights of 51 girls:

Height (in cm) Number of girls
Less than 1404
Less than 14511
Less than 15029
Less than 15540
Less than 16046
Less than 16551

Solution:

First, we convert this to a frequency distribution:

Class intervals Frequency Cumulative frequency
Below 14044
140-145711
145-1501829
150-1551140
155-160646
160-165551

Here \(n = 51\), so \(\frac{n}{2} = 25.5\)

The median class is 145-150 (cumulative frequency = 29)

Now:

  • \(l\) = 145
  • \(cf\) = 11
  • \(f\) = 18
  • \(h\) = 5

Using the formula:

\(\text{Median} = 145 + \left( \frac{25.5 - 11}{18} \right) \times 5 = 145 + \frac{72.5}{18} = 149.03\)

So, the median height of the girls is 149.03 cm.

Exercise Solutions

Exercise 13.1 - Question 1

Number of plants in 20 houses:

Number of plants 0-2 2-4 4-6 6-8 8-10 10-12 12-14
Number of houses 1 2 1 5 6 2 3

Solution:

We'll use the direct method:

Class Interval fi xi fixi
0-2111
2-4236
4-6155
6-85735
8-106954
10-1221122
12-1431339
Total 20 162

\(\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{162}{20} = 8.1\)

Therefore, the mean number of plants per house is 8.1.

I used the direct method because the values are small and calculations are straightforward.

Exercise 13.1 - Question 2

Daily wages of 50 workers:

Daily wages (in ₹) 500-520 520-540 540-560 560-580 580-600
Number of workers 12 14 8 6 10

Solution:

We'll use the step deviation method with \(a = 550\) and \(h = 20\):

Class Interval fi xi ui = (xi - 550)/20 fiui
500-52012510-2-24
520-54014530-1-14
540-560855000
560-580657016
580-60010590220
Total 50 -12

\(\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h = 550 + \left( \frac{-12}{50} \right) \times 20 = 550 - 4.8 = 545.2\)

Therefore, the mean daily wage is ₹545.20.

Exercise 13.2 - Question 1

Ages of patients admitted in a hospital:

Age (in years) 5-15 15-25 25-35 35-45 45-55 55-65
Number of patients 6 11 21 23 14 5

Solution:

Mode: The modal class is 35-45 (highest frequency = 23)

\(l = 35\), \(h = 10\), \(f_1 = 23\), \(f_0 = 21\), \(f_2 = 14\)

\(\text{Mode} = 35 + \left( \frac{23 - 21}{46 - 21 - 14} \right) \times 10 = 35 + \frac{20}{11} = 36.8\)

So, the modal age is 36.8 years.

Mean: Using direct method:

Class Interval fi xi fixi
5-1561060
15-251120220
25-352130630
35-452340920
45-551450700
55-65560300
Total 80 2830

\(\bar{x} = \frac{2830}{80} = 35.375\)

So, the mean age is 35.375 years.

The mode and mean are different because the distribution is not perfectly symmetrical.

Exercise 13.3 - Question 1

Distribution of 100 families by their expenditure:

Expenditure (in ₹) 1000-1500 1500-2000 2000-2500 2500-3000 3000-3500 3500-4000 4000-4500 4500-5000
Number of families 24 40 33 28 30 22 16 7

Solution:

First, we calculate cumulative frequencies:

Class Interval fi Cumulative frequency
1000-15002424
1500-20004064
2000-25003397
2500-300028125
3000-350030155
3500-400022177
4000-450016193
4500-50007200

Here \(n = 200\), so \(\frac{n}{2} = 100\)

The median class is 2000-2500 (cumulative frequency = 97)

Now:

  • \(l\) = 2000
  • \(cf\) = 64
  • \(f\) = 33
  • \(h\) = 500

Using the formula:

\(\text{Median} = 2000 + \left( \frac{100 - 64}{33} \right) \times 500 = 2000 + \frac{18000}{33} = 2545.45\)

So, the median expenditure is ₹2545.45.