1. A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?

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 |

Which method did you use for finding the mean, and why?

In order to find the mean value, we will use direct method because the numerical value of \(f_i \) and \( x_i \) are small.

The midpoint of the given interval is found by formula:
\(Midpoint \ (x_i) \ = \ \frac{upper \ limit \ + \ lower \ limit}{2} \)

No. of plants (Class interval) | No. of houses (Frequency \( (f_i) \) ) | Mid-point \((x_i)\) | \(f_i x_i \) |

0-2 | 1 | 1 | 1 |

2-4 | 2 | 3 | 6 |

4-6 | 1 | 5 | 5 |

6-8 | 5 | 7 | 35 |

8-10 | 6 | 9 | 54 |

10-12 | 2 | 11 | 22 |

12-14 | 3 | 13 | 39 |

\( \sum \ f_i \ = \ 20 \) | \( \sum \ f_i x_i \ = \ 162 \) |

The formula to find the mean is: \( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \ = \ \frac{162}{20} \)

\( = \ 8.1 \)

∴ The mean number of plants per house is 8.1

2. Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.

Daily wages (in Rs.) |
100-120 |
120-140 |
140-160 |
160-180 |
180-200 |

Number of workers |
12 |
14 |
8 |
6 |
10 |

The midpoint of the given interval is found by formula:
\(Midpoint \ (x_i) \ = \ \frac{upper \ limit \ + \ lower \ limit}{2} \)

In this case, the value of mid-point ( \(x_i\)) is very large,so let us assume the mean value, A = 150 and class interval is h = 20.

So, \(u_i \ = \ \frac{x_i \ - \ A}{h} \ => \ u_i \ = \ \frac{x_i \ - \ 150}{20} \)

Daily wages (Class interval) | Number of workers (frequency (\(f_i\)) ) | Mid-point ( \(x_i \) ) | \( u_i \ = \ \frac{x_i \ - \ 150}{20} \) | \(f_i u_i \) |

100-120 | 12 | 110 | -2 | -24 |

120-140 | 14 | 130 | -1 | -14 |

140-160 | 8 | 150 | 0 | 0 |

160-180 | 6 | 170 | 1 | 6 |

180-200 | 10 | 190 | 2 | 20 |

Total | \( \sum f_i \ = \ 50 \) | \( \sum f_i u_i \ = \ -12 \) |

So, the formula to find out the mean is: \( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\( = \ 150 \ + \ 20 \ × \ \frac{-12}{50} \)

\(= \ 150 \ - \ 4.8 \)

\(= \ 145.20 \)

∴ Mean daily wage of the workers = Rs. 145.20

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Daily Pocket Allowence(in Rs) |
11-13 |
13-15 |
15-17 |
17-19 |
19-21 |
21-23 |
23-35 |

Number of children |
7 |
6 |
9 |
13 |
f |
5 |
4 |

Given, \( \overline{x} \ = \ 18 \)

Class interval | Number of children (frequency (\(f_i\)) ) | Mid-point \( (x_i) \) | \(f_i u_i \) |

11-13 | 7 | 12 | 84 |

13-15 | 6 | 14 | 84 |

15-17 | 9 | 16 | 144 |

17-19 | 13 | 18 | 234 |

19-21 | f | 20 | 20f |

21-23 | 5 | 22 | 110 |

23-25 | 4 | 24 | 96 |

Total | \( \sum f_i \ = \ 44 \ + \ f \) | \( \sum f_i x_i \ = \ 752 \ + \ 20f \) |

The mean formula is\( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \ => \ 18 \ = \ \frac{752 \ + \ 20f}{44 \ + \ f} \)

\(=> \ 18(44 \ + \ f) \ = \ (752 \ + \ 20f) \ => \ 792 \ + \ 18f \ = \ 752 \ + \ 20f \)

\(=> \ 792 \ - \ 752 \ = \ 20f \ - \ 18f \ => \ 40 \ = \ 2f \)

\( => \ f \ = \ 20 \)

So, the missing frequency, f = 20.

4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Number of heartbeats per minute |
65-68 |
68-71 |
71-74 |
74-77 |
77-80 |
80-83 |
83-86 |

Number of women |
2 |
4 |
3 |
8 |
7 |
4 |
2 |

From the given data, let us assume the mean as A = 75.5

Class size (h) = 3

So, \(u_i \ = \ \frac{x_i \ - \ A}{h} \ => \ u_i \ = \ \frac{x_i \ - \ 75.5}{h} \)

Class Interval | Number of women ( \( f_i \) ) | Mid-point ( \(x_i \) ) | \(u_i \ = \ \frac{x_i \ - \ 75.5}{h} \) | \( f_i u_i \) |

65-68 | 2 | 66.5 | -3 | -6 |

68-71 | 4 | 69.5 | -2 | -8 |

71-74 | 3 | 72.5 | -1 | -3 |

74-77 | 8 | 75.5 | 0 | 0 |

77-80 | 7 | 78.5 | 1 | 7 |

80-83 | 4 | 81.5 | 3 | 8 |

83-86 | 2 | 84.5 | 3 | 6 |

Sum f_{i}= 30 |
Sum f_{i}u_{i }= 4 |

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\(= 75.5 \ + \ 3 \ × \ \frac{4}{30} \ = \ 75.5 \ + \ \frac{4}{10} \ = \ 75.9 \)

∴ The mean heart beats per minute for these women is 75.9

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Number of mangoes |
50-52 |
53-55 |
56-58 |
59-61 |
62-64 |

Number of boxes |
15 |
110 |
135 |
115 |
25 |

Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals are 1

Here, assumed mean (A) = 57, Class size (h) = 3

Here, the step deviation is used because the frequency values are big.

Class Interval | Number of boxes \( (f_i) \) | Mid-point \( (x_i) \) | \( d_i \ = \ x_i \ - \ A \) | \( f_i d_i \) |

49.5-52.5 | 15 | 51 | -6 | 90 |

52.5-55.5 | 110 | 54 | -3 | -330 |

55.5-58.5 | 135 | 57 | 0 | 0 |

58.5-61.5 | 115 | 60 | 3 | 345 |

61.5-64.5 | 25 | 63 | 6 | 150 |

\( \sum f_i \ = \ 400 \) | \( \sum f_i d_i \ = \ 75 \) |

The formula to find out the Mean is:

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i d_i}{ \sum f_i} \ = \ 57 \ + \ 3 \frac{75}{400} \)

\(= \ 57 \ + \ 0.1875 \ = \ 57.19 \)

∴ The mean number of mangoes kept in a packing box is 57.19

6. The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

Daily expenditure(in Rs) |
100-150 |
150-200 |
200-250 |
250-300 |
300-350 |

Number of households |
4 |
5 |
12 |
2 |
2 |

Let is assume the mean (A) = 225, Class size (h) = 50

Class Interval | Number of households \( ( f_i ) \) | Mid-point \( ( x_i ) \) | \( d_i \ = \ x_i \ - \ A \) | \( u_i \ = \ \frac{d_i}{50} \) | \( f_i u_i \) |

100-150 | 4 | 125 | -100 | -2 | -8 |

150-200 | 5 | 175 | -50 | -1 | -5 |

200-250 | 12 | 225 | 0 | 0 | 0 |

250-300 | 2 | 275 | 50 | 1 | 2 |

300-350 | 2 | 325 | 100 | 2 | 4 |

\( \sum f_i \ = \ 25 \) | \( \sum f_i u_i \ = \ -7 \) |

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i u_i}{ \sum f_i} \ = \ 225 \ + \ 50 \frac{-7}{25} \)

\(= \ 225 \ - \ 14 \ = \ 211 \)

∴ The mean daily expenditure on food is 211

7. To find out the concentration of SO_{2} in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO_{2} in the air.

Concentration of SO_{2} ( in ppm) |
Frequency |

0.00 – 0.04 |
4 |

0.04 – 0.08 |
9 |

0.08 – 0.12 |
9 |

0.12 – 0.16 |
2 |

0.16 – 0.20 |
4 |

0.20 – 0.24 |
2 |

Calculate mean by direct method:

Concentration of SO_{2 }(in ppm) |
Frequency \( ( f_i ) \) | Mid-point \( (x_i) \) | \( f_i x_i \) |

0.00-0.04 | 4 | 0.02 | 0.08 |

0.04-0.08 | 9 | 0.06 | 0.54 |

0.08-0.12 | 9 | 0.10 | 0.90 |

0.12-0.16 | 2 | 0.14 | 0.28 |

0.16-0.20 | 4 | 0.18 | 0.72 |

0.20-0.24 | 2 | 0.20 | 0.40 |

Total | \( \sum f_i \ = \ 30 \) | \( \sum f_i x_i \ = \ 2.96 \) |

The formula to find out the mean is

\( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \ = \ \frac{2.96}{30} \)

\( = \ 0.099 \) ppm

∴ The mean concentration of SO_{2} in air is 0.099 ppm.

8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of days |
0-6 |
6-10 |
10-14 |
14-20 |
20-28 |
28-38 |
38-40 |

Number of students |
11 |
10 |
7 |
4 |
4 |
3 |
1 |

Class interval | Frequency \( (f_i) \) | Mid-point \( (x_i) \) | \( f_i x_i \) |

0-6 | 11 | 3 | 33 |

6-10 | 10 | 8 | 80 |

10-14 | 7 | 12 | 84 |

14-20 | 4 | 17 | 68 |

20-28 | 4 | 24 | 96 |

28-38 | 3 | 33 | 99 |

38-40 | 1 | 39 | 39 |

\( \sum f_i \ = \ 40 \) | \( \sum f_i x_i \ = \ 499 \) |

The mean formula is,

\( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \ = \ \frac{499}{40} \)

\(= \ 12.48 \) days

∴ The mean number of days a student was absent = 12.48.

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %) |
45-55 |
55-65 |
65-75 |
75-85 |
85-98 |

Number of cities |
3 |
10 |
11 |
8 |
3 |

In this case, the value of mid-point (x_{i}) is very large,

so let us assume the mean value, A = 70 and class interval is h = 10.

So, \( u_i \ = \ \frac{x_i \ - \ A}{h} \)

\(=> \ u_i \ = \ \frac{x_i \ - \ 70}{10} \)

Class Interval | Frequency \( (f_i) \) | \( (x_i) \) | \( d_i \ = \ x_i \ - \ A \) | \(u_i \ = \ \frac{d_i}{h} \) | \( f_i u_i \) |

45-55 | 3 | 50 | -20 | -2 | -6 |

55-65 | 10 | 60 | -10 | -1 | -10 |

65-75 | 11 | 70 | 0 | 0 | 0 |

75-85 | 8 | 80 | 10 | 1 | 8 |

85-95 | 3 | 90 | 20 | 2 | 6 |

\( \sum f_i \ = \ 35 \) | \( \sum f_i u_i \ = \ -2 \) |

So, \( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \ = \ 70 \ + \ \frac{-2}{35} \ ? \ 10 \)

\(= \ 69.42 \)

∴ The mean literacy part is 69.42

1. The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Age (in years) |
5-15 |
15-25 |
25-35 |
35-45 |
45-55 |
55-65 |

Number of patients |
6 |
11 |
21 |
23 |
14 |
5 |

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

The class 35-45 has the maximum frequency, therefore, this is the modal class.

Here l = 35, h = 10, f_{1} = 23, f_{0} = 21, f_{2} =14

Now, let us substitute these values in the formula

\(Mode \ = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 35 \ + \ \frac{23 \ - \ 21}{2 \ × \ 23 \ - \ 21 \ - \ 14} \ × \ 10 \)

\(= \ 35 \ + \ \frac{2}{46 \ - \ 21 \ - \ 14} \ × \ 10 \)

\(= \ 35 \ + \ \frac{2}{11} \ × \ 10 \)

\(= \ 35 \ + \ 1.8 \ = \ 36.8 \)

Now calculate Mean

Class Interval | Frequency ( \(f_i\) ) | Mid-point ( \(x_i\) ) | \( f_i x_i \) |

5-15 | 6 | 10 | 60 |

15-25 | 11 | 20 | 220 |

25-35 | 21 | 30 | 630 |

35-45 | 23 | 40 | 920 |

45-55 | 14 | 50 | 700 |

55-65 | 5 | 60 | 300 |

\( \sum f_i \ = \ 80 \) | \( \sum f_i x_i \ = \ 2830 \) |

The mean formula is

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

\(= \ \frac{2830}{80} \ = \ 35.37 \) years

∴ The mean of the given data = 35.37 years

2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components.

Lifetime (in hours) |
0-20 |
20-40 |
40-60 |
60-80 |
80-100 |
100-120 |

Frequency |
10 |
35 |
52 |
61 |
38 |
29 |

Determine the modal lifetimes of the components.

The class 60-80 has the maximum frequency, therefore, this is the modal class.

Here l = 60 , h = 20, f_{1} = 61, f_{0} = 52 and f_{2} = 38 .

Now, let us substitute these values in the formula

\(Mode \ = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \ = \ 60 \ + \ ( \frac{61 \ - \ 52}{2 \ × \ 61 \ - \ 52 \ - \ 38}) \ × \ 20 \)

\(= \ 60 \ + \ ( \frac{61 \ - \ 52}{122 \ - \ 52 \ - \ 38}) \ × \ 20 \)

\(= \ 60 \ + \ \frac{9}{32} \ × \ 20 \)

\( = \ 60 \ + \ 5.625 \ = \ 65.625 \)

∴ The modal lifetimes of the components is 65.625 hours.

3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

Expenditure |
Number of families |

1000-1500 |
24 |

1500-2000 |
40 |

2000-2500 |
33 |

2500-3000 |
28 |

3000-3500 |
30 |

3500-4000 |
22 |

4000-4500 |
16 |

4500-5000 |
7 |

The class 1500 – 2000 has the maximum frequency,therefore, this is the modal class.

Here l = 1500, h = 500, f_{1} =40,f_{0} =24 and f_{2} =33

Now, let us substitute these values in the formula

Mode \(= \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 1500 \ + \ ( \frac{40 \ - \ 24}{2 \ × \ 40 \ - \ 24 \ - \ 33}) \ × \ 500\)

\(= \ 1500 \ + \ \frac{40 \ -\ 24}{80 \ ? \ 24 \ ? \ 33} \ × \ 500 \)

\(= \ 1500 \ + \ \frac{16}{23} \ × \ 500 \)

\(= \ 1500 \ + \ 347.83 \ = \ 1847.83 \)

∴ Modal monthly expenditure = Rs. 1847.83

Let the assumed mean be A = 2750 and h = 500.

Class Interval | \(f_i \) | \( x_i \) | \(d_i \ = \ x_i \ - \ A \) | \(u_i \ = \ \frac{d_i}{h} \) | \(f_i u_i \) |

1000-1500 | 24 | 1250 | -1500 | -3 | -72 |

1500-2000 | 40 | 1750 | -1000 | -2 | -80 |

2000-2500 | 33 | 2250 | -500 | -1 | -33 |

2500-3000 | 28 | 2750 | 0 | 0 | 0 |

3000-3500 | 30 | 3250 | 500 | 1 | 30 |

3500-4000 | 22 | 3750 | 1000 | 2 | 44 |

4000-4500 | 16 | 4250 | 1500 | 3 | 48 |

4500-5000 | 7 | 4750 | 2000 | 4 | 28 |

\( \sum f_i \ = \ 200 \) | \( \sum f_i u_i \ = \ -35 \) |

The formula to calculate the mean,

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\(= \ 2750 \ + \ ( \frac{-35}{200}) \ × \ 500 \)

\( = \ 2750 \ - \ 87.50 \ = \ 2662.50 \)

So, the mean monthly expenditure of the families is Rs 2662.50

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures

No of Students per teacher |
Number of states / U.T |

15-20 |
3 |

20-25 |
8 |

25-30 |
9 |

30-35 |
10 |

35-40 |
3 |

40-45 |
0 |

45-50 |
0 |

50-55 |
2 |

The class 30-35 has the maximum frequency, therefore, this is the modal class.

Here l = 30, h = 5, f_{1} = 10, f_{0} = 9 and f_{2} = 3

Now, let us substitute these values in the formula

Mode \(= \ l \ + \ ( \frac{f_1 \ - \ f0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\( = \ 30 \ + \ ( \frac{10 \ - \ 9}{2 \ × \ 10 \ - \ 9 \ - \ 3}) \ × \ 5 \)

\(= \ 30 \ + \ \frac{10 \ - \ 9}{20 \ - \ 9 \ - \ 3} \ × \ 5 \)

\(= \ 30 \ + \ \frac{1}{8} \ × \ 5 \ = \ 30 \ + \ 0.625 \)

\( = \ 30.625 \ = \ 30.6 \)

Calculation of mean :

Class Interval | Frequency \( (f_i) \) | Mid-point \( (x_i) \) | \( f_i x_i \) |

15-20 | 3 | 17.5 | 52.5 |

20-25 | 8 | 22.5 | 180.0 |

25-30 | 9 | 27.5 | 247.5 |

30-35 | 10 | 32.5 | 325.0 |

35-40 | 3 | 37.5 | 112.5 |

40-45 | 0 | 42.5 | 0 |

45-50 | 0 | 47.5 | 0 |

50-55 | 2 | 52.5 | 105.5 |

\( \sum f_i \ = \ 35 \) | \( \sum f_i x_i \ = \ 1022.5 \) |

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

\(= \ \frac{1022.5}{35} \ = \ 29.2 \)

∴ mean = 29.2

5. The given distribution shows the number of runs scored by some top batsmen of the world in one- day international cricket matches.

Find the mode of the data.

Run Scored |
Number of Batsman |

3000-4000 |
4 |

4000-5000 |
18 |

5000-6000 |
9 |

6000-7000 |
7 |

7000-8000 |
6 |

8000-9000 |
3 |

9000-10000 |
1 |

10000-11000 |
1 |

The class 4000-5000 has the maximum frequency ,

therefore, this is the modal class.

Here l = 4000, h = 1000, f_{1} = 18, f_{0} = 4 and f_{2} = 9

Now, let us substitute these values in the formula

Mode \(= l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 4000 \ + \ ( \frac{18 \ - \ 4}{2 \ × \ 18 \ - \ 4 \ - \ 9}) \ × \ 1000 \)

\(= \ 4000 \ + \ \frac{18 \ - \ 4}{36 \ - \ 4 \ - \ 9} \ × \ 1000 \)

\(= \ 4000 \ + \ \frac{14}{23} \ × \ 1000 \)

\(= \ 4000 \ + \ 608.695 \ = 4608.7 \)

∴ The mode of the given data is 4608.7 runs.

6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:

Number of cars |
Frequency |

0-10 |
7 |

10-20 |
14 |

20-30 |
13 |

30-40 |
12 |

40-50 |
20 |

50-60 |
11 |

60-70 |
15 |

70-80 |
8 |

The class 40-50 has the maximum frequency, therefore, this is the modal class.

Here l = 40, h =10, f_{1} = 20, f_{0} = 12 and f_{2} = 11

Now, let us substitute these values in the formula

Mode \( = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 40 \ + \ ( \frac{20 \ - \ 12}{40 \ - \ 12 \ - \ 11}) \ × \ 10 \)

\( = \ 40 \ + \ \frac{8}{17} \ × \ 10 \)

\( = \ 40+4.7 \ = \ 44.7 \)

∴ Mode = 44.7 cars

1. The following frequency distribution gives the monthly consumption of an electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption(in units) |
No. of customers |

65-85 |
4 |

85-105 |
5 |

105-125 |
13 |

125-145 |
20 |

145-165 |
14 |

165-185 |
8 |

185-205 |
4 |

Class Interval |
Frequency |
Cumulative frequency |

65-85 | 4 | 4 |

85-105 | 5 | 9 |

105-125 | 13 | 22 |

125-145 | 20 | 42 |

145-165 | 14 | 56 |

165-185 | 8 | 64 |

185-205 | 4 | 68 |

N=68 |

We have n = 68 \( => \ \frac{n}{2} \ = \ \frac{68}{2} \ = \ 34 \)

The cumulative frequency just greater than \( \frac{n}{2} \) , is 42 and the corresponding class is 125-145.

Thus, 125-145 is the median class such that \( \frac{n}{2} \ = \ 34 \), l = 125, c.f. = 22, f = 20, h = 20 .

Median \(= \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f}) \ × \ h \),

Median \( = \ 125 \ + \ ( \frac{34 \ - \ 22}{20}) \ × \ 20 \ = \ 125 \ + \ 12 \ = \ 137 \) units

Calculation of mean

Let the assumed mean, A = 135, class interval, h = 20

so \(u_i \ = \ \frac{x_i \ - \ A}{h} \ = \ \frac{x_i \ - \ 135}{20} \)

Class Interval |
\( f_i \) |
\( x_i \) |
\( d_i \ = \ x_i \ - \ A \) |
\( u_i \ = \ \frac{d_i}{h} \) |
\( f_i u_i \) |

65-85 | 4 | 75 | -60 | -3 | -12 |

85-105 | 5 | 95 | -40 | -2 | -10 |

105-125 | 13 | 115 | -20 | -1 | -13 |

125-145 | 20 | 135 | 0 | 0 | 0 |

145-165 | 14 | 155 | 20 | 1 | 14 |

165-185 | 8 | 175 | 40 | 2 | 16 |

185-205 | 4 | 195 | 60 | 3 | 12 |

\( \sum f_i \ = \ 68 \) | \( \sum f_i u_i \ = \ 7 \) |

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \ = \ 135 \ + \ 20( \frac{7}{68}) \)

Mean=137.05

In this case, mean, median and mode are more/less equal in this distribution.

2. If the median of a distribution given below is 28.5 then, find the value of an x &y.

Class Interval |
Frequency |

0-10 |
5 |

10-20 |
x |

20-30 |
20 |

30-40 |
15 |

40-50 |
y |

50-60 |
5 |

Total |
60 |

Given that median is 28.5 and \(n \ = \ \sum f_i \ = \ 60 \)

Here n = 60 \( => \ \frac{n}{2} \ = \ 30 \)

Since the median is given to be 28.5, thus the median class is 20-30.

∴ l = 20, h = 10, f = 20 and cf = 5 + x

∴ median \(= \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

\(=> \ 28.5 \ = \ 20 \ + \ \frac{30 \ - \ 5 \ - \ x}{20} \ × \ 10 \)

\( => \ 57 \ = \ 40 \ + \ 25 \ - \ x \ => \ x \ = \ 65 \ - \ 57 \ => \ x \ = \ 8 \)

Also, 45 + x + y = 60 [ As n = 60 ]

\(=> \ 45 \ + \ 8 \ + \ y \ = \ 60 \)

\(=> \ y \ = \ 60 \ - \ 53 \ => \ y \ = \ 7 \)

∴ x = 8, y = 7

Q.3 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 onwards but less than 60 year.

Age(in years) |
Number of policy holder |

Below 20 |
2 |

Below 25 |
6 |

Below 30 |
24 |

Below 35 |
45 |

Below 40 |
78 |

Below 45 |
89 |

Below 50 |
92 |

Below 55 |
98 |

Below 60 |
100 |

We are given the cumulative frequency distribution. So, we first construct a frequency table from the given cumulative frequency distribution and then we will make necessary computations to compute median .

Class interval | Frequency | Cumulative frequency |

15-20 | 2 | 2 |

20-25 | 4 | 6 |

25-30 | 18 | 24 |

30-35 | 21 | 45 |

35-40 | 33 | 78 |

40-45 | 11 | 89 |

45-50 | 3 | 92 |

50-55 | 6 | 98 |

55-60 | 2 | 100 |

Here \( n \ = \ \sum f_i \ = \ 100 \ => \ \frac{n}{2} \ = \ 50 \)

We see that the cumulative frequency just greater than \( \frac{n}{2} \), i.e., 50 is 78 and the corresponding class is 35-40. So, 35-40 is the median class.

∴ \( \frac{n}{2} \ = \ 50 \) , l = 35, c.f = 45 , f = 33, h = 5

Now, let us substitute these values in the formula

Median \( = \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

\(= \ 35 \ + \ \frac{50 \ - \ 45}{33} \ × \ 5 \ = \ 35 \ + \ \frac{5}{33} \ × \ 5 \ = \ 35 \ + \ 0.76 \ = \ 35.76 \)

∴ The median age = 35.76 years

Q.4 The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

Find the median length of leaves.

Length(in mm) |
Number of leaves |

118-126 |
3 |

127-135 |
5 |

136-144 |
9 |

145-153 |
12 |

154-162 |
5 |

163-171 |
4 |

172-180 |
2 |

Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.

Class Interval | Frequency | Cumulative frequency |

117.5-126.5 | 3 | 3 |

126.5-135.5 | 5 | 8 |

135.5-144.5 | 9 | 17 |

144.5-153.5 | 12 | 29 |

153.5-162.5 | 5 | 34 |

162.5-171.5 | 4 | 38 |

171.5-180.5 | 2 | 40 |

So, n = 40 and \( \frac{n}{2} \ = \ 20 \)

Median class = 144.5-153.5

then, l = 144.5, cf = 17, f = 12, h = 9

Now we know that, Median \(= \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

=> Median \( = \ 144.5 \ + \ ( \frac{20 \ - \ 17}{12}) \ × \ 9 \ = \ 144.5 \ + \ \frac{3}{12} \ × \ 9 \)

\( = \ 144.5 \ + \ 2.25 \ = \ 146.75 \)

∴ The median length of the leaves = 146.75 mm.

Q.5 The following table gives the distribution of the life time of 400 neon lamps:

Find the median lifetime of a lamp.

Lifetime(in hours) |
Number of lamps |

1500-2000 |
14 |

2000-2500 |
56 |

2500-3000 |
60 |

3000-3500 |
86 |

3500-4000 |
74 |

4000-4500 |
62 |

4500-5000 |
48 |

Class Interval |
Frequency |
Cumulative |

1500-2000 | 14 | 14 |

2000-2500 | 56 | 70 |

2500-3000 | 60 | 130 |

3000-3500 | 86 | 216 |

3500-4000 | 74 | 290 |

4000-4500 | 62 | 352 |

4500-5000 | 48 | 400 |

n = 400 => \( \frac{n}{2} \ = \ 200 \)

Median class = 3000-3500

∴ l = 3000, cf = 130, f = 86, h = 500

Median \( = \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

Median \(= \ 3000 \ + \ ( \frac{200 \ - \ 130}{86}) \ × \ 500 \)

\(= \ 3000 \ + \ \frac{70}{86} \ × \ 500 \)

\(= \ 3000 \ + \ 406.98 \)

\(= \ 3406.98 \)

∴ median lifetime of lamps = 3406.98 hours

6. In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:

Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.

Number of letters |
1-4 |
4-7 |
7-10 |
10-13 |
13-16 |
16-19 |

Number of surnames |
6 |
30 |
40 |
16 |
4 |
4 |

Class Interval |
Frequency |
Cumulative Frequency |

1-4 | 6 | 6 |

4-7 | 30 | 36 |

7-10 | 40 | 76 |

10-13 | 16 | 92 |

13-16 | 4 | 96 |

16-19 | 4 | 100 |

We have n = 100 => \( \frac{n}{2} \ = \ 50 \)

The cumulative frequency just greater than \( \frac{n}{2} \) is 76 and the corresponding class is 7-10 . Thus 7-10 is the median class such that

\( \frac{n}{2} \ = \ 50 \), l = 7, cf = 36, f = 40 and h = 3

Median \( = \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

\( = \ 7 \ + \ ( \frac{50 \ - \ 36}{40}) \ × \ 3 \ = \ 7 \ + \ \frac{14}{40} \ × \ 3 \)

\(= \ 7 \ + \ 1.05 \ = \ 8.05 \)

Class Interval |
\( f_i \) |
\( x_i \) |
\( f_i x_i \) |

1-4 | 6 | 2.5 | 15 |

4-7 | 30 | 5.5 | 165 |

7-10 | 40 | 8.5 | 340 |

10-13 | 16 | 11.5 | 184 |

13-16 | 4 | 14.5 | 51 |

16-19 | 4 | 17.5 | 70 |

\( \sum f_i \ = \ 100 \) | \( \sum f_i x_i \ = \ 825 \) |

∴ Mean = \( \overline{x} \ = \ \frac{ \sum f_i x_i}{ \sum f_i} \ = \ \frac{832}{100} \ = \ 8.32 \)

Calculation of mode

The class 7-10 has the maximum frequency therefore, this is the modal class.

Here l = 7 , h = 3, f

Mode \( = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 7 \ + \ \frac{40 \ - \ 30}{80 \ - \ 30 \ - \ 16} \ × \ 3 \)

\(= \ 7 \ + \ \frac{10}{34} \ × \ 3 \ = \ 7 \ + \ 0.88 \ = \ 7.88 \)

∴ median = 8.05, mean = 8.32 and mode = 7.88

Q.7 The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight(in kg) |
40-45 |
45-50 |
50-55 |
55-60 |
60-65 |
65-70 |
70-75 |

Number of students |
2 |
3 |
8 |
6 |
6 |
3 |
2 |

Class Interval |
Frequency |
Cumulative frequency |

40-45 | 2 | 2 |

45-50 | 3 | 5 |

50-55 | 8 | 13 |

55-60 | 6 | 19 |

60-65 | 6 | 25 |

65-70 | 3 | 28 |

70-75 | 2 | 30 |

We have n = 30 \( => \ \frac{n}{2} \ = \ 15 \)

The cumulative frequency just greater than n2 is 19 and the corresponding class is 55-60.

Thus, 55-60 is the median class such that n2=15, l = 55, f = 6, cf = 13 and h = 5

Median \(= \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

\(= \ 55 \ + \ ( \frac{15 \ - \ 13}{6}) \ × \ 5 \)

\(= \ 55 \ + \ \frac{2}{6} \ × \ 5 \ = \ 55 \ + \ 1.67 \ = \ 56.67 \)

∴ The median weight = 56.67 kg .

1. The following distribution gives the daily income of 50 workers of a factory.

Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.

Daily income in Rupees |
100-120 |
120-140 |
140-160 |
160-180 |
180-200 |

Number of workers |
12 |
14 |
8 |
6 |
10 |

Daily income |
Frequency |
Cumulative Frequency |

Less than 120 | 12 | 12 |

Less than 140 | 14 | 26 |

Less than 160 | 8 | 34 |

Less than 180 | 6 | 40 |

Less than 200 | 10 | 50 |

From the table plot the points corresponding to the ordered pairs such as (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and the plotted points are joined to get a smooth curve and the obtained curve is known as less than type ogive curve

2. During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight in kg |
Number of students |

Less than 38 |
0 |

Less than 40 |
3 |

Less than 42 |
5 |

Less than 44 |
9 |

Less than 46 |
14 |

Less than 48 |
28 |

Less than 50 |
32 |

Less than 52 |
35 |

Draw a less than type ogive for the given data. Hence obtain the median weight from
the graph and verify the result by using the formula

From the given data, to represent the table in the form of graph, choose the upper limits of the class intervals are in x-axis and frequencies on y-axis by choosing the convenient scale.

Now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper an join them to get a smooth curve. The curve obtained is known as less than type ogive.

Locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From the point, draw a perpendicular line to the x-axis.
The intersection point perpendicular to x-axis is the median of the given data.
Now, to find the mode by making a table.

Class interval | Number of students(Frequency) | Cumulative Frequency |

Less than 38 | 0 | 0 |

Less than 40 | 3-0=3 | 3 |

Less than 42 | 5-3=2 | 8 |

Less than 44 | 9-5=4 | 9 |

Less than 46 | 14-9=5 | 14 |

Less than 48 | 28-14=14 | 28 |

Less than 50 | 32-28=4 | 32 |

Less than 52 | 35-22=3 | 35 |

The class 46-48 has the maximum frequency, therefore, this is modal class

Here, l = 46, h = 2, f

Now, Mode \( = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 46 \ + \ \frac{14 \ - \ 5}{28 \ - \ 5 \ - \ 4} \ × \ 2 \ = \ 46 \ + \ \frac{9}{19} \ × \ 2 \)

\(= \ 46 \ + \ 0.95 \ = \ 46.95 \)

Thus, mode is verified.

3. The following tables gives production yield per hectare of wheat of 100 farms of a village.

Change the distribution to a more than type distribution and draw its ogive.

Production Yield |
50-55 |
55-60 |
60-65 |
65-70 |
70-75 |
75-80 |

Number of farms |
2 |
8 |
12 |
24 |
38 |
16 |

Converting the given distribution to a more than type distribution, we get

Production Yield (kg/ha) | Number of farms |

More than or equal to 50 | 100 |

More than or equal to 55 | 100-2 = 98 |

More than or equal to 60 | 98-8= 90 |

More than or equal to 65 | 90-12=78 |

More than or equal to 70 | 78-24=54 |

More than or equal to 75 | 54-38 =16 |