CBSE Class 10 Mathematics Statistics VBQs Set 09

Question. A survey was conducted by a group of students as a part of their environment awareness programme, 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.
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?

Answer:
Sol.
Number of plants | Class mark (\( x_i \)) | Number of houses (\( f_i \)) | \( f_i x_i \)
0 – 2 | 1 | 1 | 01
2 – 4 | 3 | 2 | 06
4 – 6 | 5 | 1 | 05
6 – 8 | 7 | 5 | 35
8 – 10 | 9 | 6 | 54
10 – 12 | 11 | 2 | 22
12 – 14 | 13 | 3 | 39
Total | \( \sum f_i = 20 \) | | \( \sum f_i x_i = 162 \)
We have, Mean \( (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} = \frac{162}{20} = 8.1 \) plants
The mean of the data is 8.1.
Since, the values of \( x_i \) and \( f_i \) are small, so we have used direct method to find the mean.

Question. Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in Rs.): 500 – 520, 520 – 540, 540 – 560, 560 – 580, 580 – 600
Number of workers: 12, 14, 8, 6, 10
Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer:
Sol. Here \( h = 20 \)
Daily wages (in Rs.) | Class mark (\( x_i \)) | Number of workers (\( f_i \)) | \( d_i = x_i - 550 \) | \( u_i = \frac{d_i}{20} \) | \( f_i u_i \)
500 – 520 | 510 | 12 | –40 | –2 | –24
520 – 540 | 530 | 14 | –20 | –1 | –14
540 – 560 | 550 = a (Let) | 8 | 0 | 0 | 0
560 – 580 | 570 | 6 | 20 | 1 | 6
580 – 600 | 590 | 10 | 40 | 2 | 20
Total | | \( \sum f_i = 50 \) | | | \( \sum f_i u_i = -12 \)
We have, Mean \( = a + \frac{\sum f_i u_i}{\sum f_i} \times h = 550 + \left( \frac{-12}{50} \right) \times 20 = 550 - \frac{24}{5} = \frac{2750 - 24}{5} = \frac{2726}{5} = \) Rs. 545.20

Question. 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 allowance (in Rs.): 11–13, 13–15, 15–17, 17–19, 19–21, 21–23, 23–25
Number of children: 7, 6, 9, 13, f, 5, 4

Answer:
Sol.
Daily pocket allowance (in Rs.) | Class mark (\( x_i \)) | Number of children (\( f_i \)) | \( d_i = x_i - 18 \) | \( f_i d_i \)
11 – 13 | 12 | 7 | –6 | –42
13 – 15 | 14 | 6 | –4 | –24
15 – 17 | 16 | 9 | –2 | –18
17 – 19 | 18 = a (Let) | 13 | 0 | 0
19 – 21 | 20 | f | 2 | 2f
21 – 23 | 22 | 5 | 4 | 20
23 – 25 | 24 | 4 | 6 | 24
Total | | \( \sum f_i = 44 + f \) | | \( \sum f_i d_i = 2f - 40 \)
We have,
Mean \( = a + \frac{\sum f_i d_i}{\sum f_i} \)

\( \implies \) \( 18 = 18 + \frac{2f - 40}{44 + f} \)

\( \implies \) \( 0 = \frac{2f - 40}{44 + f} \)

\( \implies \) \( 2f - 40 = 0 \)

\( \implies \) \( 2f = 40 \)

\( \implies \) \( f = \frac{40}{2} = 20 \)

Question. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats 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

Answer:
Sol.
Number of heart beats per minute | Class mark (\( x_i \)) | Number of women (\( f_i \)) | \( u_i = \frac{x_i - a}{3} \) | \( f_i u_i \)
65 – 68 | 66.5 | 2 | –3 | –6
68 – 71 | 69.5 | 4 | –2 | –8
71 – 74 | 72.5 | 3 | –1 | –3
74 – 77 | 75.5 = a | 8 | 0 | 0
77 – 80 | 78.5 | 7 | 1 | 7
80 – 83 | 81.5 | 4 | 2 | 8
83 – 86 | 84.5 | 2 | 3 | 6
Total | | \( \sum f_i = 30 \) | | \( \sum f_i u_i = 4 \)
We have, Mean \( = a + \frac{\sum f_i u_i}{\sum f_i} \times h \)
\( = 75.5 + \frac{4 \times 3}{30} = 75.5 + 0.4 = 75.9 \)

Question. 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.
Number of mangoes: 50–52, 53–55, 56–58, 59–61, 62–64
Number of boxes: 15, 110, 135, 115, 25
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Answer:
Sol. Here \( h = 3 \)
Number of mangoes | Class mark (\( x_i \)) | Number of boxes (\( f_i \)) | \( u_i = \frac{x_i - 57}{3} \) | \( f_i u_i \)
50 – 52 | 51 | 15 | –2 | –30
53 – 55 | 54 | 110 | –1 | –110
56 – 58 | 57 = a (Let) | 135 | 0 | 0
59 – 61 | 60 | 115 | 1 | 115
62 – 64 | 63 | 25 | 2 | 50
Total | | \( \sum f_i = 400 \) | | \( \sum f_i u_i = 25 \)
We have, Mean \( = a + \frac{\sum f_i u_i}{\sum f_i} \times h \)
\( = 57 + \frac{25 \times 3}{400} = 57 + 0.19 = 57.19 \) mangoes
Step deviation method.

Question. The table below shows the daily expenditure on food of 25 households in a locality.
Daily expenditure (in Rs.): 100–150, 150–200, 200–250, 250–300, 300–350
Number of households: 4, 5, 12, 2, 2
Find the mean daily expenditure on food by a suitable method.

Answer:
Sol. Here \( h = 50 \)
Daily expenditure (in Rs.) | Classmark (\( x_i \)) | Number of households (\( f_i \)) | \( u_i = \frac{x_i - 225}{50} \) | \( f_i u_i \)
100 – 150 | 125 | 4 | –2 | –8
150 – 200 | 175 | 5 | –1 | –5
200 – 250 | 225 = a (Let) | 12 | 0 | 0
250 – 300 | 275 | 2 | 1 | 2
300 – 350 | 325 | 2 | 2 | 4
Total | | \( \sum f_i = 25 \) | | \( \sum f_i u_i = -7 \)
We have, Mean \( = a + \frac{\sum f_i u_i}{\sum f_i} \times h \)
\( = 225 + \frac{-7}{25} \times 50 = 225 - 14 = \) Rs. 211

Question. 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:
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
Find the mean concentration of \( SO_2 \) in the air.

Answer:
Sol. Here \( h = 0.04 \)
Concentration of \( SO_2 \) (in ppm) | Class mark (\( x_i \)) | Frequency (\( f_i \)) | \( u_i = \frac{x_i - 0.10}{0.04} \) | \( f_i u_i \)
0.00 – 0.04 | 0.02 | 4 | –2 | –8
0.04 – 0.08 | 0.06 | 9 | –1 | –9
0.08 – 0.12 | 0.10 = a (Let) | 9 | 0 | 0
0.12 – 0.16 | 0.14 | 2 | 1 | 2
0.16 – 0.20 | 0.18 | 4 | 2 | 8
0.20 – 0.24 | 0.22 | 2 | 3 | 6
Total | | \( \sum f_i = 30 \) | | \( \sum f_i u_i = -1 \)
We have, Mean \( = a + \frac{\sum f_i u_i}{\sum f_i} \times h \)
\( = 0.10 + \frac{(-1) \times 0.04}{30} = 0.10 - 0.001 = 0.099 \) ppm.

Question. 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

Answer:
Sol.
Number of days | Class mark (\( x_i \)) | Number of students (\( f_i \)) | \( d_i = x_i - 17 \) | \( f_i d_i \)
0 – 6 | 3 | 11 | –14 | –154
6 – 10 | 8 | 10 | –9 | –90
10 – 14 | 12 | 7 | –5 | –35
14 – 20 | 17 = a (Let) | 4 | 0 | 0
20 – 28 | 24 | 4 | 7 | 28
28 – 38 | 33 | 3 | 16 | 48
38 – 40 | 39 | 1 | 22 | 22
Total | | \( \sum f_i = 40 \) | | \( \sum f_i d_i = -181 \)
We have, Mean \( = a + \frac{\sum f_i d_i}{\sum f_i} = 17 + \frac{-181}{40} = 17 - 4.52 = 12.48 \) days

Question. 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 – 95
Number of cities: 3, 10, 11, 8, 3

Answer:
Sol. Here \( h = 10 \)
Literacy rate (in%) | Class mark (\( x_i \)) | Number of cities (\( f_i \)) | \( u_i = \frac{x_i - 70}{10} \) | \( f_i u_i \)
45 – 55 | 50 | 3 | – 2 | – 6
55 – 65 | 60 | 10 | – 1 | – 10
65 – 75 | 70 = a (Let) | 11 | 0 | 0
75 – 85 | 80 | 8 | 1 | 8
85 – 95 | 90 | 3 | 2 | 6
Total | | \( \sum f_i = 35 \) | | \( \sum f_i u_i = - 2 \)
Mean \( = a + \frac{\sum f_i u_i}{\sum f_i} \times h = 70 + \frac{(- 2) \times 10}{35} = 70 - 0.57 = 69.43\% \)

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

Answer:
Sol. For Mode:
Age (in years) | 5–15 | 15–25 | 25–35 | 35–45 | 45–55 | 55–65
Number of patients | 6 | 11 | 21 | 23 | 14 | 5
Maximum frequency = 23
\( \therefore \) Modal class = 35 – 45
Here, \( l = 35, f_1 = 23, f_0 = 21, f_2 = 14, h = 10 \)
Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 35 + \left( \frac{23 - 21}{46 - 21 - 14} \right) \times 10 = 35 + \frac{2}{11} \times 10 = 35 + \frac{20}{11} = 36.8 \) years
For Mean:
Age (in years) | Class mark (\( x_i \)) | Number of patients (\( f_i \)) | \( u_i = \frac{x_i - 30}{10} \) | \( f_i u_i \)
5 – 15 | 10 | 6 | –2 | –12
15 – 25 | 20 | 11 | –1 | –11
25 – 35 | 30 = a (Let) | 21 | 0 | 0
35 – 45 | 40 | 23 | 1 | 23
45 – 55 | 50 | 14 | 2 | 28
55 – 65 | 60 | 5 | 3 | 15
Total | | \( \sum f_i = 80 \) | | \( \sum f_i u_i = 43 \)
Here, \( a = 30, \sum f_i u_i = 43, \sum f_i = 80, h = 10 \)
We have, Mean \( = a + \frac{\sum f_i u_i}{\sum f_i} \times h = 30 + \frac{43 \times 10}{80} = 30 + 5.37 = 35.37 \) years
We conclude that the maximum number of patients in the hospital are of the age 36.8 years. While on an average, the age of patient admitted to the hospital is 35.37 years.

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

Answer:
Sol. Lifetimes (in hours) | 0–20 | 20–40 | 40–60 | 60–80 | 80–100 | 100–120
Frequency | 10 | 35 | 52 | 61 | 38 | 29
Here, Maximum frequency = 61
\( \therefore \) Modal class = 60 – 80
Here, \( l = 60, f_0 = 52, f_1 = 61, f_2 = 38, h = 20 \)
We have, Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 60 + \left( \frac{61 - 52}{122 - 52 - 38} \right) \times 20 = 60 + \frac{9 \times 20}{32} = 60 + \frac{45}{8} = 60 + 5.625 = 65.625 \) hours

Question. 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 (in Rs.): 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

Answer:
Sol. Here, maximum frequency = 40
\( \therefore \) Modal class = 1500 – 2000 and \( l = 1500, f_0 = 24, f_1 = 40, f_2 = 33 \)
Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 1500 + \left( \frac{40 - 24}{80 - 24 - 33} \right) \times 500 = 1500 + \frac{16}{23} \times 500 = 1500 + 347.83 = \) Rs. 1847.83
For Mean:
Expenditure (in Rs.) | Class mark (\( x_i \)) | Number of families (\( f_i \)) | \( u_i = \frac{x_i - 2750}{500} \) | \( f_i u_i \)
1000 – 1500 | 1250 | 24 | –3 | –72
1500 – 2000 | 1750 | 40 | –2 | –80
2000 – 2500 | 2250 | 33 | –1 | –33
2500 – 3000 | 2750 = a (Let) | 28 | 0 | 0
3000 – 3500 | 3250 | 30 | 1 | 30
3500 – 4000 | 3750 | 22 | 2 | 44
4000 – 4500 | 4250 | 16 | 3 | 48
4500 – 5000 | 4750 | 7 | 4 | 28
Total | | \( \sum f_i = 200 \) | | \( \sum f_i u_i = -35 \)
Here, \( a = 2750, \sum f_i = 200, \sum f_i u_i = -35, h = 500 \)
\( \therefore \) Mean \( = a + \frac{\sum f_i u_i}{\sum f_i} \times h = 2750 + \left( \frac{-35}{200} \right) \times 500 = 2750 - \frac{175}{2} = 2750 - 87.50 = \) Rs. 2662.50

Question. 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.
Number of students per teacher: 15 – 20, 20 – 25, 25 – 30, 30 – 35, 35 – 40, 40 – 45, 45 – 50, 50 – 55
Number of states /U.T.: 3, 8, 9, 10, 3, 0, 0, 2

Answer:
Sol.
Number of students per teacher | Number of states/U.T. (\( f_i \)) | Class mark (\( x_i \)) | \( u_i = \frac{x_i - 32.5}{5} \) | \( f_i u_i \)
15 – 20 | 3 | 17.5 | –3 | –9
20 – 25 | 8 | 22.5 | –2 | –16
25 – 30 | 9 | 27.5 | –1 | –9
30 – 35 | 10 | 32.5 = a (Let) | 0 | 0
35 – 40 | 3 | 37.5 | 1 | 3
40 – 45 | 0 | 42.5 | 2 | 0
45 – 50 | 0 | 47.5 | 3 | 0
50 – 55 | 2 | 52.5 | 4 | 8
Total | \( \sum f_i = 35 \) | | | \( \sum f_i u_i = -23 \)
Here, \( a = 32.5, \sum f_i u_i = -23, h = 5, \sum f_i = 35 \)
We have, Mean \( = a + \frac{\sum f_i u_i}{\sum f_i} \times h = 32.5 + \left( \frac{-23}{35} \right) \times 5 = 32.5 - \frac{23}{7} = 32.5 - 3.3 = 29.2 \)
Maximum frequency = 10; \( \therefore \) Modal class = 30 – 35 Here, \( l = 30, f_0 = 9, f_1 = 10, f_2 = 3 \)
Also we have, Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 30 + \left( \frac{10 - 9}{20 - 9 - 3} \right) \times 5 = 30 + \frac{5}{8} = 30.6 \)
Hence, we conclude that most states/U.T. have a student teacher ratio of 30.6 and on an average this ratio is 29.2.

Question. The given distribution shows the number of runs scored by some top batsmen of the world in one – day international cricket matches.
Runs scored: 3000 – 4000, 4000 – 5000, 5000 – 6000, 6000 – 7000, 7000 – 8000, 8000 – 9000, 9000 – 10000, 10000 – 11000
Number of batsmen: 4, 18, 9, 7, 6, 3, 1, 1
Find the mode of the data.

Answer:
Sol. Runs scored | Number of batsmen (\( f_i \))
3000 – 4000 | 4
4000 – 5000 | 18
5000 – 6000 | 9
6000 – 7000 | 7
7000 – 8000 | 6
8000 – 9000 | 3
9000 – 10000 | 1
10000 – 11000 | 1
Maximum frequency = 18,
\( \therefore \) Modal class = 4000 – 5000; Here, \( l = 4000, f_0 = 4, f_1 = 18, f_2 = 9, h = 1000 \)
Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 4000 + \left( \frac{18 - 4}{36 - 4 - 9} \right) \times 1000 = 4000 + \frac{14000}{23} = 4000 + 608.7 = 4608.7 \) runs

Question. 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: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60, 60–70, 70–80
Frequency: 7, 14, 13, 12, 20, 11, 15, 8

Answer:
Sol.
Number of cars | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 | 60–70 | 70–80
Frequency (\( f_i \)) | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8
Maximum frequency = 20
\( \therefore \) Modal class = 40 – 50;
Here, \( l = 40, f_0 = 12, f_1 = 20, f_2 = 11, h = 10 \)
\( \therefore \) Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 40 + \left( \frac{20 - 12}{40 - 12 - 11} \right) \times 10 = 40 + 4.7 = 44.7 \) cars

Question. 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.
Monthly consumption (in units): 65 – 85, 85 – 105, 105 – 125, 125 – 145, 145 – 165, 165 – 185, 185 – 205
Number of consumers: 4, 5, 13, 20, 14, 8, 4

Answer:
Sol.
Monthly consumption (in units) | Number of consumers (\( f_i \)) | Cumulative frequency (\( cf \)) | Class mark (\( x_i \)) | \( u_i = \frac{x_i - a}{h} \) | \( f_i u_i \)
65 – 85 | 4 | 4 | 75 | –3 | –12
85 – 105 | 5 | 9 | 95 | –2 | –10
105 – 125 | 13 | 22 | 115 | –1 | –13
125 – 145 | 20 | 42 | 135 = a (Let) | 0 | 0
145 – 165 | 14 | 56 | 155 | 1 | 14
165 – 185 | 8 | 64 | 175 | 2 | 16
185 – 205 | 4 | 68 | 195 | 3 | 12
Total | \( \sum f_i = 68 \) | | | | \( \sum f_i u_i = 7 \)

Question. If the median of the distribution given below is 28.5, find the values of \( x \) and \( y \).
Class interval | Frequency
0 – 10 | 5
10 – 20 | \( x \)
20 – 30 | 20
30 – 40 | 15
40 – 50 | \( y \)
50 – 60 | 5
Total | 60

Answer: Given, Median \( = 28.5 \) which lies in the class (20 – 30),
Here, \( l = 20, f = 20, cf = 5 + x, h = 10, n = 60 \)
Class interval | Frequency | Cumulative
0 – 10 | 5 | 5
10 – 20 | \( x \) | \( 5 + x \)
20 – 30 | 20 | \( 25 + x \)
30 – 40 | 15 | \( 40 + x \)
40 – 50 | \( y \) | \( 40 + x + y \)
50 – 60 | 5 | \( 45 + x + y \)
Total | 60 |
Median \( = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \)
\( = 20 + \left( \frac{30 - (5 + x)}{20} \right) \times 10 \)
\( = 20 + \frac{30 - 5 - x}{2} \)

\( \implies \) \( 28.5 = 20 + \frac{25 - x}{2} \)

\( \implies \) \( 28.5 - 20 = \frac{25 - x}{2} \)

\( \implies \) \( 8.5 \times 2 = 25 - x \)

\( \implies \) \( x = 25 - 17 \)

\( \implies \) \( x = 8 \)
Again \( 45 + x + y = 60 \)

\( \implies \) \( x + y = 60 - 45 = 15 \)
\( 8 + y = 15 \)

\( \implies \) \( y = 15 - 8 = 7 \)

Question. 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 holders
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

Answer: Age (in years) | Number of policy holders | Cumulative frequency
0 – 20 | 2 | 2
20 – 25 | 6 – 2 = 4 | 6
25 – 30 | 24 – 6 = 18 | 24
30 – 35 | 45 – 24 = 21 | 45
35 – 40 | 78 – 45 = 33 | 78
40 – 45 | 89 – 78 = 11 | 89
45 – 50 | 92 – 89 = 3 | 92
50 – 55 | 98 – 92 = 6 | 98
55 – 60 | 100 – 98 = 2 | 100
Total | 100 |
Here, \( \frac{n}{2} = \frac{100}{2} = 50 \)

\( \therefore \) Median class \( = 35 – 40 \), So, \( l = 35, cf = 45, h = 5, f = 33 \)
We have, Median \( = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \)
\( = 35 + \left( \frac{50 - 45}{33} \right) \times 5 = 35 + \frac{25}{33} = 35 + 0.76 = 35.76 \) years

Question. The lengths of 40 leaves of a plant are measured correct to nearest millimetre, and the data obtained is represented in the following table:
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
Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, …, 171.5 – 180.5.)

Answer: Length (in mm) | Number of leaves | \( cf \)
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
Total | 40 |
Here, \( \frac{n}{2} = \frac{40}{2} = 20 \)
So, Median class\( = 144.5 – 153.5 \)
Here, \( l = 144.5, f = 12, cf = 17, h = 9 \)
We have, Median \( = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \)
\( = 144.5 + \left( \frac{20 - 17}{12} \right) \times 9 \)
\( = 144.5 + \frac{9}{4} = 146.75 \) mm

Question. The following table gives the distribution of the life time of 400 neon lamps:
Life time (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
Find the median life time of a lamp.

Answer: Life time (in hours) | Number of lamps | \( cf \)
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
Total | 400 |
Here, \( \frac{n}{2} = \frac{400}{2} = 200 \)

\( \therefore \) Median class\( = 3000 – 3500 \);
So, \( f = 86, cf = 130, h = 500, l = 3000 \)
We have, Median \( = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h = 3000 + \left( \frac{200 - 130}{86} \right) \times 500 = 3000 + \frac{35000}{86} \)
\( = 3000 + 406.98 = 3406.98 \) hours

Question. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
Number of letters: 1 – 4, 4 – 7, 7 – 10, 10 – 13, 13 – 16, 16 – 19
Number of surnames: 6, 30, 40, 16, 4, 4
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.

Answer: Number of letters | Number of surnames | \( cf \)
1 – 4 | 6 | 6
4 – 7 | 30 | 36
7 – 10 | 40 | 76
10 – 13 | 16 | 92
13 – 16 | 4 | 96
16 – 19 | 4 | 100
Total | 100 |
Here, \( \frac{n}{2} = \frac{100}{2} = 50 \)

\( \therefore \) Median class \( = 7 – 10 \);
So, \( l = 7, f = 40, cf = 36, h = 3 \)
We have, Median \( = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h = 7 + \left( \frac{50 - 36}{40} \right) \times 3 = 7 + \frac{42}{40} = 8.05 \)
Here, Modal class \( = 7 – 10 \)
So, \( l = 7, f_0 = 30, f_1 = 40, f_2 = 16, h = 3 \)

\( \therefore \) Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 7 + \left( \frac{40 - 30}{80 - 30 - 16} \right) \times 3 = 7 + \frac{30}{34} = 7.88 \)
For mean:
Number of letters | Frequency (\( f_i \)) | Class mark (\( x_i \)) | \( u_i = \frac{x_i - 8.5}{3} \) | \( f_i u_i \)
1 – 4 | 6 | 2.5 | –2 | –12
4 – 7 | 30 | 5.5 | –1 | –30
7 – 10 | 40 | 8.5 = a (Let) | 0 | 0
10 – 13 | 16 | 11.5 | 1 | 16
13 – 16 | 4 | 14.5 | 2 | 8
16 – 19 | 4 | 17.5 | 3 | 12
Total | \( \sum f_i = 100 \) | | | \( \sum f_i u_i = - 6 \)
We have, Mean \( = a + \frac{\sum f_i u_i}{\sum f_i} \times h = 8.5 + \frac{- 6}{100} \times 3 = 8.5 - 0.18 = 8.32 \)

Question. The distribution below gives the weight 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

Answer: Weight (in kg) | Number of students (\( f_i \)) | \( cf \)
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
Total | 30 |
Here, \( \frac{n}{2} = \frac{30}{2} = 15 \),

\( \therefore \) Median class \( = 55 – 60 \),
So, \( l = 55, f = 6, cf = 13, h = 5 \)
Median weight \( = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h = 55 + \left( \frac{15 - 13}{6} \right) \times 5 = 55 + \frac{5}{3} = 55 + 1.67 = 56.67 \) kg

Question. Choose the correct answer from the given four options : In the formula \( \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \) for finding the mean of grouped data \( d_i \)’s are deviation from a of
(a) lower limits of the classes
(b) upper limits of the classes
(c) mid-points of the classes
(d) frequencies of the class marks

Answer: (c) mid-points of the classes.

Question. While computing mean of grouped data, we assume that the frequencies are
(a) evenly distributed over all the classes
(b) centred at the classmarks of the classes
(c) centred at the upper limits of the classes
(d) centred at the lower limits of the classes

Answer: (b) centred at the classmarks of the classes.

Question. In the formula \( \bar{x} = a + h \left( \frac{\sum f_i u_i}{\sum f_i} \right) \), for finding the mean of grouped frequency distribution, \( u_i = \)
(a) \( \frac{x_i + a}{h} \)
(b) \( h(x_i - a) \)
(c) \( \frac{x_i - a}{h} \)
(d) \( \frac{a - x_i}{h} \)

Answer: (c) \( \frac{x_i - a}{h} \)

Question. The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its
(a) mean
(b) median
(c) mode
(d) All of the options

Answer: (b) median

Question. Find the mean marks of students for the following distribution:
Marks | Number of students
0 and above | 80
10 and above | 77
20 and above | 72
30 and above | 65
40 and above | 55
50 and above | 43
60 and above | 28
70 and above | 16
80 and above | 10
90 and above | 8
100 and above | 0

Answer: By step deviation method
Let \( a = 55 \) and \( h = 10 \)
Class | Frequency (\( f_i \)) | Mid-value (\( x_i \)) | \( u_i = \frac{x_i - a}{h} \) | \( u_i f_i \)
0 – 10 | 3 | 5 | – 5 | – 15
10 – 20 | 5 | 15 | – 4 | – 20
20 – 30 | 7 | 25 | – 3 | – 21
30 – 40 | 10 | 35 | – 2 | – 20
40 – 50 | 12 | 45 | – 1 | – 12
50 – 60 | 15 | 55 | 0 | 0
60 – 70 | 12 | 65 | 1 | 12
70 – 80 | 6 | 75 | 2 | 12
80 – 90 | 2 | 85 | 3 | 6
90 – 100 | 8 | 95 | 4 | 32
Total | \( \sum f_i = 80 \) | | | \( \sum f_i u_i = - 26 \)
Mean \( (\bar{x}) = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h = 55 + \left( \frac{-26}{80} \right) \times 10 = 55 - 3.25 = 51.75 \)

Question. Determine the mean of the following distribution:
Marks | Number of students
Below 10 | 5
Below 20 | 9
Below 30 | 17
Below 40 | 29
Below 50 | 45
Below 60 | 60
Below 70 | 70
Below 80 | 78
Below 90 | 83
Below 100 | 85

Answer: \( a = 45, h = 10 \).
Class | Frequency (\( f_i \)) | Mid-value (\( x_i \)) | \( u_i = \frac{x_i - a}{h} \) | \( f_i u_i \)
0 – 10 | 5 | 5 | – 4 | – 20
10 – 20 | 4 | 15 | – 3 | – 12
20 – 30 | 8 | 25 | – 2 | – 16
30 – 40 | 12 | 35 | – 1 | – 12
40 – 50 | 16 | 45 | 0 | 0
50 – 60 | 15 | 55 | 1 | 15
60 – 70 | 10 | 65 | 2 | 20
70 – 80 | 8 | 75 | 3 | 24
80 – 90 | 5 | 85 | 4 | 20
90 – 100 | 2 | 95 | 5 | 10
Total | \( \sum f_i = 85 \) | | | \( \sum f_i u_i = 29 \)
\( \bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h = 45 + \left( \frac{29}{85} \right) \times 10 = 45 + 3.41 = 48.41 \)

Question. Form the frequency distribution table from the following data:
Marks (out of 90) | Number of candidates
More than or equal to 80 | 4
More than or equal to 70 | 6
More than or equal to 60 | 11
More than or equal to 50 | 17
More than or equal to 40 | 23
More than or equal to 30 | 27
More than or equal to 20 | 30
More than or equal to 10 | 32
More than or equal to 0 | 34

Answer: Frequency distribution table is as follows:
Marks | No. of candidates
0 – 10 | 2
10 – 20 | 2
20 – 30 | 3
30 – 40 | 4
40 – 50 | 6
50 – 60 | 6
60 – 70 | 5
70 – 80 | 2
80 – 90 | 4

Question. The following table shows the cumulative frequency distribution of marks of 800 students in an examination:
Marks | Below 10 | Below 20 | Below 30 | Below 40 | Below 50 | Below 60 | Below 70 | Below 80 | Below 90 | Below 100
No. of students | 10 | 50 | 130 | 270 | 440 | 570 | 670 | 740 | 780 | 800
Construct a frequency distribution table for the data above.

Answer: Frequency distribution table:
Marks | No. of students
0 – 10 | 10
10 – 20 | 40
20 – 30 | 80
30 – 40 | 140
40 – 50 | 170
50 – 60 | 130
60 – 70 | 100
70 – 80 | 70
80 – 90 | 40
90 – 100 | 20