CBSE Class 10 Mathematics Statistics VBQs Set G

Question. The mean and median of 100 observations are 50 and 52 respectively. The value of the largest observation is 100. It was later found that it is 110 not 100. Find the true mean and median.
Answer: Mean = \( \frac{\sum fx}{\sum f} \)
\( \Rightarrow 50 = \frac{\sum fx}{100} \)
\( \Rightarrow \sum fx = 5000 \)
Correct, \( \sum fx' = 5000 - 100 + 110 = 5010 \)
\( \therefore \) Correct Mean = \( \frac{5010}{100} = 50.1 \)
Median will remain same i.e., median = 52

Question. The data regarding marks obtained by 48 students of a class in a class test is given below:
Marks obtained: 0 – 5, 5 – 10, 10 – 15, 15 – 20, 20 – 25, 25 – 30, 30 – 35, 35 – 40, 40 – 45, 45 – 50
Number of students: 1, 0, 2, 0, 0, 10, 25, 7, 2, 1
Calculate the modal marks of students.
Answer: Modal class is 30 – 35, \( l = 30, f_1 = 25, f_0 = 10, f_2 = 7 \) and \( h = 5 \)
Mode = \( l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( \Rightarrow \) Mode = \( 30 + \left( \frac{25 - 10}{50 - 10 - 7} \right) \times 5 \)
= \( 30 + 2.27 \) or 32.27 approx.

Question. Given below is the distribution of weekly pocket money received by students of a class. Calculate the pocket money that is received by most of the students.
Pocket money (in ₹): 0 – 20, 20 – 40, 40 – 60, 60 – 80, 80 – 100, 100 – 120, 120 – 140
Number of students: 2, 2, 3, 12, 18, 5, 2
Answer: Here, Modal Class = 80 – 100
\( \therefore l = 80, f_1 = 18, f_2 = 5, f_0 = 12 \) and \( h = 20 \)
\( \therefore \) Mode = \( l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
= \( 80 + \left( \frac{18 - 12}{36 - 12 - 5} \right) \times 20 \)
= \( 80 + \frac{6}{19} \times 20 \)
= \( 80 + 6.31 = 86.31 \) (approx.)
Hence, mode = 86.31.

Question. The mean of the following frequency distribution is 25. Find the value of p.
Class interval: 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50
Frequency: 4, 6, 10, 6, p
Answer:
Class-Interval | Mid-Point \( x_i \) | \( f_i \) | \( f_i x_i \)
0 – 10 | 5 | 4 | 20
10 – 20 | 15 | 6 | 90
20 – 30 | 25 | 10 | 250
30 – 40 | 35 | 6 | 210
40 – 50 | 45 | p | 45p
Total | | \( \sum f_i = 26 + p \) | \( \sum f_i x_i = 570 + 45p \)
Mean, \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \)
\( \Rightarrow 25 = \frac{570 + 45p}{26 + p} \)
\( \Rightarrow 650 + 25p = 570 + 45p \)
\( \Rightarrow 650 - 570 = 45p - 25p \)
\( \therefore p = 4 \) (Approx.)

Question. Given below is a frequency distribution table showing daily income of 100 workers of a factory:
Daily income of workers (in ₹): 200 – 300, 300 – 400, 400 – 500, 500 – 600, 600 – 700
Number of workers: 12, 18, 35, 20, 15
Convert this table to a cumulative frequency distribution table of 'more than type'.
Answer:
Daily income of workers (in ₹) | Number of workers
More than 200 | 100
More than 300 | 88
More than 400 | 70
More than 500 | 35
More than 600 | 15
More than 700 | 0

Question. The following are the ages of 300 patients getting medical treatment in a hospital on a particular day:
Age (in years): 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60, 60 – 70
Number of patients: 60, 42, 55, 70, 53, 20
Form the “less than type” cumulative frequency distribution table.
Answer:
Age (in years) | Number of Patients
Less than 20 | 60
Less than 30 | 102
Less than 40 | 157
Less than 50 | 227
Less than 60 | 280
Less than 70 | 300

Short Answer Type Questions-

Question. The median of the following data is 16. Find the missing frequencies a and b, if the total of the frequencies is 70.
Class: 0 – 5, 5 – 10, 10 – 15, 15 – 20, 20 – 25, 25 – 30, 30 – 35, 35 – 40
Frequency: 12, a, 12, 15, b, 6, 6, 4
Answer:
Class | Frequency (f) | Cumulative frequency (c.f.)
0 – 5 | 12 | 12
5 – 10 | a | 12 + a
10 – 15 | 12 | 24 + a
15 – 20 | 15 | 39 + a
20 – 25 | b | 39 + a + b
25 – 30 | 6 | 45 + a + b
30 – 35 | 6 | 51 + a + b
35 – 40 | 4 | 55 + a + b
Total | 70 |
According to question, \( 55 + a + b = 70 \)
\( a + b = 15 \) ...(i)
Median = \( l + \left( \frac{\frac{n}{2} - c.f.}{f} \right) \times h \)
\( 16 = 15 + \frac{35 - (24 + a)}{15} \times 5 \)
\( 1 = \frac{11 - a}{3} \)
\( a = 8 \)
Substituting the value of a in equation (i), we get \( 8 + b = 15 \Rightarrow b = 7 \)

Question. The mode of the following data is 67. Find the missing frequency x.
Class: 40 – 50, 50 – 60, 60 – 70, 70 – 80, 80 – 90
Frequency: 5, x, 15, 12, 7
Answer: Modal class is the class having the maximum frequency, i.e., Modal class = 60 – 70
Then, \( l = 60, f_1 = 15, f_0 = x, f_2 = 12 \) and \( h = 10 \)
\( \therefore Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( 67 = 60 + \left( \frac{15 - x}{30 - x - 12} \right) \times 10 \)
\( 7 = \frac{15 - x}{18 - x} \times 10 \)
\( 7 \times (18 - x) = 10(15 - x) \)
\( 126 - 7x = 150 - 10x \)
\( 3x = 24 \Rightarrow x = 8 \)

Question. Find the mode of the following frequency distribution.
Class: 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60, 60 – 70
Frequency: 8, 10, 10, 16, 12, 6, 7
Answer: Modal-class = 30 – 40
\( \therefore l = 30, f_0 = 10, f_1 = 16, f_2 = 12, h = 10 \)
Mode = \( l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
= \( 30 + \left( \frac{16 - 10}{2 \times 16 - 10 - 12} \right) \times 10 \)
= \( 30 + \left( \frac{6}{32 - 22} \right) \times 10 \)
= \( 30 + \frac{6}{10} \times 10 = 36 \)

Question. The mean of the following distribution is 53. Find the missing frequency k?
Class: 0 – 20, 20 – 40, 40 – 60, 60 – 80, 80 – 100
Frequency: 12, 15, 32, k, 13
Answer:
Class Interval | Frequency \( (f_i) \) | Class Marks \( (x_i) \) | \( f_i x_i \)
0 – 20 | 12 | 10 | 120
20 – 40 | 15 | 30 | 450
40 – 60 | 32 | 50 | 1600
60 – 80 | k | 70 | 70k
80 – 100 | 13 | 90 | 1170
Total | \( \sum f_i = 72 + k \) | | \( \sum f_i x_i = 3340 + 70k \)
Given, Mean = 53
Mean = \( \frac{\sum f_i x_i}{\sum f_i} \)
\( \Rightarrow 53 = \frac{3340 + 70k}{72 + k} \)
\( \Rightarrow 53(72 + k) = 3340 + 70k \)
\( \Rightarrow 3816 + 53k = 3340 + 70k \)
\( \Rightarrow 17k = 476 \Rightarrow k = 28 \)

Question. The table below show the salaries of 280 persons:
Salary (in thousand ₹): 5 – 10, 10 – 15, 15 – 20, 20 – 25, 25 – 30, 30 – 35, 35 – 40, 40 – 45, 45 – 50
No. of Persons: 49, 133, 63, 15, 6, 7, 4, 2, 1
Calculate the median salary of the data.
Answer:
Salary (in thousand ₹) | No. of Persons | c.f.
5 – 10 | 49 | 49
10 – 15 | 133 = f | 182
15 – 20 | 63 | 245
20 – 25 | 15 | 260
25 – 30 | 6 | 266
30 – 35 | 7 | 273
35 – 40 | 4 | 277
40 – 45 | 2 | 279
45 – 50 | 1 | 280
\( \frac{n}{2} = \frac{280}{2} = 140 \). Median class = 10 – 15.
Median = \( l + \left( \frac{\frac{n}{2} - c.f.}{f} \right) \times h \)
= \( 10 + \left( \frac{140 - 49}{133} \right) \times 5 \)
= \( 10 + \frac{5 \times 91}{133} = 13.42 \)
Hence, median salary is ₹ 13.42 thousand or ₹ 13420 (approx).

Question. If the mean of the following data is 14.7, find the values of p and q.
Class: 0 – 6, 6 – 12, 12 – 18, 18 – 24, 24 – 30, 30 – 36, 36 – 42 | Total
Frequency: 10, p, 4, 7, q, 4, 1 | 40
Answer:
Class | \( x_i \) | \( f_i \) | \( x_i f_i \)
0 – 6 | 3 | 10 | 30
6 – 12 | 9 | p | 9p
12 – 18 | 15 | 4 | 60
18 – 24 | 21 | 7 | 147
24 – 30 | 27 | q | 27q
30 – 36 | 33 | 4 | 132
36 – 42 | 39 | 1 | 39
Total | | \( \sum f_i = 26 + p + q = 40 \) | \( \sum x_i f_i = 408 + 9p + 27q \)
Given, \( \sum f_i = 40 \Rightarrow p + q = 14 \) ...(i)
Mean, \( \bar{x} = \frac{\sum x_i f_i}{\sum f_i} \)
\( 14.7 = \frac{408 + 9p + 27q}{40} \)
\( \Rightarrow p + 3q = 20 \) ...(ii)
Subtracting (i) from (ii), \( 2q = 6 \Rightarrow q = 3 \)
Putting q in (i), \( p = 14 - 3 = 11 \)
Hence, \( p = 11, q = 3 \)

Question. Find the median of the following data:
Height (in cm): Less than 120, Less than 140, Less than 160, Less than 180, Less than 200
Number of students: 12, 26, 34, 40, 50
Answer:
Height (in cm) | Frequency | c.f.
less than 120 | 12 | 12
120 – 140 | 14 | 26
140 – 160 | 8 | 34
160 – 180 | 6 | 40
180 – 200 | 10 | 50
Total \( n = 50 \)
\( Median = \frac{n}{2} = \frac{50}{2} = 25 \)
Median Class = 120 – 140
Median = \( l + \left( \frac{\frac{n}{2} - c.f.}{f} \right) \times h \)
= \( 120 + \left( \frac{25 - 12}{14} \right) \times 20 \)
= \( 120 + \frac{260}{14} = 120 + 18.57 = 138.57 \)

Question. Form the frequency distribution table from the following data:
Marks (out of 90) | Number of students (c.f.)
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:
Marks (out of 90) | Number of students (c.f.) | C.I. | Number of students \( (f_i) \)
More than or equal to 0 | 34 | 0 – 10 | 2
More than or equal to 10 | 32 | 10 – 20 | 2
More than or equal to 20 | 30 | 20 – 30 | 3
More than or equal to 30 | 27 | 30 – 40 | 4
More than or equal to 40 | 23 | 40 – 50 | 6
More than or equal to 50 | 17 | 50 – 60 | 6
More than or equal to 60 | 11 | 60 – 70 | 5
More than or equal to 70 | 6 | 70 – 80 | 2
More than or equal to 80 | 4 | 80 – 90 | 4

Long Answer Type Questions

Question. The median of the following data is 525. Find the values of x and y, if total frequency is 100.
Class: 0 – 100, 100 – 200, 200 – 300, 300 – 400, 400 – 500, 500 – 600, 600 – 700, 700 – 800, 800 – 900, 900 – 1000
Frequency: 2, 5, x, 12, 17, 20, y, 9, 7, 4
Answer:
Class Interval | Frequency | Cumulative frequency
0 – 100 | 2 | 2
100 – 200 | 5 | 7
200 – 300 | x | 7 + x
300 – 400 | 12 | 19 + x
400 – 500 | 17 | 36 + x
500 – 600 | 20 | 56 + x
600 – 700 | y | 56 + x + y
700 – 800 | 9 | 65 + x + y
800 – 900 | 7 | 72 + x + y
900 – 1000 | 4 | 76 + x + y
Total \( n = 100 \)
Also, \( 76 + x + y = 100 \Rightarrow x + y = 24 \) ...(i)
Given, Median = 525, which lies between class 500 – 600. Median class = 500 – 600.
Median = \( l + \left( \frac{\frac{n}{2} - c.f.}{f} \right) \times h \)
\( 525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100 \)
\( 25 = (50 - 36 - x) \times 5 \)
\( 5 = 14 - x \Rightarrow x = 9 \)
Putting the value of x in (i), we get \( y = 24 - 9 = 15 \)
Hence, \( x = 9 \) and \( y = 15 \).

Question. The mean of the following distribution is 18. Find the frequency \( f \) of the class 19 – 21.
Class: 11 – 13, 13 – 15, 15 – 17, 17 – 19, 19 – 21, 21 – 23, 23 – 25
Frequency: 3, 6, 9, 13, f, 5, 4
Answer:
Sol.
Class | Class mark (\(x\)) | Frequency (\(f\)) | \(fx\)
11 – 13 | 12 | 3 | 36
13 – 15 | 14 | 6 | 84
15 – 17 | 16 | 9 | 144
17 – 19 | 18 | 13 | 234
19 – 21 | 20 | \(f\) | \(20f\)
21 – 23 | 22 | 5 | 110
23 – 25 | 24 | 4 | 96
Total | | \(\sum f = 40 + f\) | \(\sum fx = 704 + 20f\)

Mean = \( 18 = \frac{704 + 20f}{40 + f} \)
\( \Rightarrow 720 + 18f = 704 + 20f \)
\( \Rightarrow 16 = 2f \)
\( \Rightarrow f = 8 \)

Question. Daily wages of 110 workers, obtained in a survey, are tabulated below:
Daily Wages (in ₹): 100 – 120, 120 – 140, 140 – 160, 160 – 180, 180 – 200, 200 – 220, 220 – 240
Number of Workers: 10, 15, 20, 22, 18, 12, 13
Compute the mean daily wages and modal daily wages of these workers.
Answer:
Sol. Calculation of mean:
Daily Wages (in ₹) | Number of Workers (\(f_i\)) | \(x_i\) | \(u_i\) | \(f_i u_i\)
100 – 120 | 10 | 110 | –3 | – 30
120 – 140 | 15 | 130 | –2 | – 30
140 – 160 | 20 | 150 | –1 | – 20
160 – 180 | 22 | 170 = A | 0 | 0
180 – 200 | 18 | 190 | 1 | 18
200 – 220 | 12 | 210 | 2 | 24
220 – 240 | 13 | 230 | 3 | 39
Total | \(\sum f_i = 110\) | | | \(\sum f_i u_i = 1\)

Mean daily wages = \( 170 + \frac{1}{110} \times 20 = Rs 170.19 \) (approx..)

Calculation of mode:
Here, maximum frequency, \( f_0 = 22 \).
So, corresponding class 160 – 180 is modal class.
\( l = \) lower boundary of modal class = 160
\( f_0 = \) maximum frequency = 22
\( f_1 = \) frequency of pre-modal class = 20
\( f_2 = \) frequency of post modal class = 18
\( h = \) width of modal class = \( 180 – 160 = 20 \)
Mode = \( l + \left[ \frac{f_0 - f_1}{2f_0 - f_1 - f_2} \right] h \)
Mode = \( 160 + \left[ \frac{22 - 20}{2(22) - 20 - 18} \right] \times 20 \)
= \( 160 + \frac{2}{6} \times 20 \)
= \( 160 + 6.67 = 166.67 \)
Therefore, modal wages of workers is ₹ 166.67. (Approx.)

Question. If the median of the following frequency distribution is 32.5. Find the values of \( f_1 \) and \( f_2 \).
Class: 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60, 60 – 70 | Total
Frequency: \(f_1\), 5, 9, 12, \(f_2\), 3, 2 | 40
Answer:
Sol.
Class | Frequency | Cumulative frequency
0 – 10 | \(f_1\) | \(f_1\)
10 – 20 | 5 | \(f_1 + 5\)
20 – 30 | 9 | \(f_1 + 14\)
30 – 40 | 12 | \(f_1 + 26\)
40 – 50 | \(f_2\) | \(f_1 + f_2 + 26\)
50 – 60 | 3 | \(f_1 + f_2 + 29\)
60 – 70 | 2 | \(f_1 + f_2 + 31\)

\( n = \sum f = 40 \)
\( f_1 + f_2 + 31 = 40 \)
\( \Rightarrow f_1 + f_2 = 9 \)
\( \Rightarrow f_2 = 9 – f_1 \) ...(i)
Given that median is 32.5, which lies in 30 – 40.
Hence, median class = 30 – 40.
Here; \( l = 30, \frac{n}{2} = \frac{40}{2} = 20, f = 12 \) and \( c.f. = 14 + f_1 \)
Now, median = \( l + \left( \frac{\frac{n}{2} - c.f.}{f} \right) \times h \)
\( \Rightarrow 32.5 = 30 + \left[ \frac{20 - (14 + f_1)}{12} \right] \times 10 \)
\( \Rightarrow 2.5 = \left( \frac{6 - f_1}{12} \right) \times 10 \)
\( \Rightarrow 2.5 = \frac{60 - 10f_1}{12} \)
\( \Rightarrow 30 = 60 - 10f_1 \)
\( \Rightarrow 10f_1 = 30 \)
\( \Rightarrow f_1 = 3 \)
From eq (i), we get \( f_2 = 9 – 3 = 6 \)
Hence, \( f_1 = 3 \) and \( f_2 = 6 \)

Question. Monthly expenditures on milk in 100 families of a housing society are given in the following frequency distribution :
Monthly expenditure (in ₹): 0 – 175, 175 – 350, 350 – 525, 525 – 700, 700 – 875, 875 – 1050, 1050 – 1225
Number of families: 10, 14, 15, 21, 28, 7, 5
Find the mode and median for this distribution.
Answer:
Sol.
C.I. | \(f\) | \(c.f.\)
0 – 175 | 10 | 10
175 – 350 | 14 | 24
350 – 525 | 15 | 39
525 – 700 | 21 | 60
700 – 875 | 28 | 88
875 – 1050 | 7 | 95
1050 – 1225 | 5 | 100

\( n = 100 \)
Median = \( \frac{n}{2} = \frac{100}{2} = 50 \)
Here Median class = 525 – 700
Now \( l = 525, c.f. = 39, f = 21 \) and \( h = 175 \)
Median = \( l + \left( \frac{\frac{n}{2} - c.f.}{f} \right) \times h \)
= \( 525 + \left( \frac{50 - 39}{21} \right) \times 175 \)
= \( 525 + \frac{11}{21} \times 175 \)
= \( 525 + 91.6 \)
= 616.6 (Approx.)

and Modal class = 700 – 875
Mode = \( l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
Here, \( l = 700, f_0 = 21, f_1 = 28, f_2 = 7 \) and \( h = 175 \)
Mode = \( 700 + \left( \frac{28 - 21}{2 \times 28 - 21 - 7} \right) \times 175 \)
= \( 700 + \frac{7}{28} \times 175 \)
= \( 700 + 43.75 \)
= 743.75