CBSE Class 10 Maths HOTs Statistics Set 09

Multiple-Choice Questions

Question. A grouped data is shown below:
Class Interval | Frequency | Cumulative Frequency
0 – 10 | 4 | 4
10 – 20 | \( a \) | \( 4 + a \)
20 – 30 | 4 | \( 8 + a \)
30 – 40 | \( b \) | \( 8 + a + b \)
40 – 50 | 5 | \( 13 + a + b \)
If the median of the grouped data is 22.50 and the total frequency is 20, then what is the value of \( a \)?
(a) 1
(b) 2
(c) 4
(d) 5
Answer: (d) 5
Sol. (d) By the formula of median, we get \( a = 5 \)

Question. If mean = (3 median – mode) . \( k \), then the value of \( k \) is
(a) 1
(b) 2
(c) \( \frac{1}{2} \)
(d) \( \frac{3}{2} \)
Answer: (c) \( \frac{1}{2} \)
Sol. (c) \( \because \) Mode \( = 3 \text{ median} - 2 \text{ mean} \)

\( \implies \) \( 2 \text{ mean} = 3 \text{ median} - \text{ mode} \)

\( \implies \) \( \text{mean} = \frac{1}{2}(3 \text{ median} - \text{ mode}) \)

\( \implies \) \( k = \frac{1}{2} \)

Question. If the mode of a data is 18 and the mean is 24, then median is
(a) 23
(b) 24
(c) 22
(d) 21
Answer: (c) 22
Sol. (c) \( 3 \text{ Median} = \text{Mode} + 2 \text{ Mean} = 18 + 2(24) = 18 + 48 = 66 \)
Median \( = \frac{66}{3} = 22. \)

Question. Construction of cumulative frequency table is useful in determining the
(a) mean
(b) median
(c) mode
(d) None of the options
Answer: (b) median
Sol. (b) Median

Question. Find the median for the above frequency distribution.
\( x \) | \( f \) | \( c.f. \)
1 | 8 | 8
2 | 10 | 18
3 | 11 | 29
4 | 16 | 45
5 | 20 | 65
6 | 25 | 90
7 | 15 | 105
8 | 9 | 114
9 | 6 | 120
(a) 4
(b) 5
(c) 6
(d) 7
Answer: (b) 5
Sol. (b) \( n = 120 \)

\( \implies \frac{n}{2} = 60 \)
\( \therefore \) Median is average of 60th and 61st observation

\( \implies \) Median \( = \frac{5 + 5}{2} = 5 \)

Question. Find the frequency of the class 50 – 60 based on the following:
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
(a) 12
(b) 10
(c) 15
(d) 718
Answer: (c) 15
Sol. (c) Marks | Frequency
0 – 10 | 3
10 – 20 | 5
20 – 30 | 7
30 – 40 | 10
40 – 50 | 12
50 – 60 | 15
60 – 70 | 12
70 – 80 | 6
80 – 90 | 2
90 – 100 | 8
\( \therefore \) Frequency of class 50 – 60 is 15.

Very Short Answer Type Questions 

Question. The daily wages (in Rs.) of 100 workers in a factory are given below:
Daily wages (in Rs.) | Number of workers
125 | 6
130 | 20
135 | 24
140 | 28
145 | 15
150 | 4
160 | 2
180 | 1
Find the median wage of a worker for the above data.
Answer: Sol. Wages (\( x_i \)) | No. of workers (\( f_i \)) | Cumulative frequency
125 | 6 | 6
130 | 20 | 26
135 | 24 | 50
140 | 28 | 78
145 | 15 | 93
150 | 4 | 97
160 | 2 | 99
180 | 1 | 100
Total = 100
\( n = 100 \) (even)
\( \therefore \) Median is the mean of \( \left(\frac{100}{2}\right) \)-th and \( \left(\frac{100}{2} + 1\right) \)-th observations, i.e., 50th and 51th observations.
\( \therefore \) Median \( = \frac{135 + 140}{2} = 137.5 \)
\( \therefore \) Median wage of a worker in the factory is Rs. 137.50.

Question. Find the median marks for the following distribution:
Classes | Number of Students
0 – 10 | 2
10 – 20 | 12
20 – 30 | 22
30 – 40 | 8
40 – 50 | 6
Answer: Sol. Classes | Number of students | \( c.f. \)
0 – 10 | 2 | 2
10 – 20 | 12 | 14
20 – 30 | 22 | 36
30 – 40 | 8 | 44
40 – 50 | 6 | 50
\( n = 50, \frac{n}{2} = \frac{50}{2} = 25 \), Median Class = 20 – 30
\( l = 20, f = 22, c.f. = 14, h = 10 \)
Median \( = l + \left( \frac{\frac{n}{2} - c.f.}{f} \right) \times h \)
\( = 20 + \left( \frac{25 - 14}{22} \right) \times 10 \)
\( = 20 + \frac{11}{22} \times 10 = 20 + 5 = 25 \)

Short Answer Type Questions 

Question. The table below shows the salaries of 280 persons. 
Salary (In thousand Rs.) | No. of Persons
5 – 10 | 49
10 – 15 | 133
15 – 20 | 63
20 – 25 | 15
25 – 30 | 6
30 – 35 | 7
35 – 40 | 4
40 – 45 | 2
45 – 50 | 1
Calculate the median salary of the data.
Answer: Sol. Distribution of frequencies:
Salary in thousand Rs.: 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
To find: median.
no. of people \( = 280 \).

\( \implies \frac{n}{2} = 140 \), the 140th term lies in class interval 10-15.

\( \implies \) median class = 10-15.
\( l=10, h=5, f=133, \frac{n}{2}=140, cf=49 \).
We know, median \( = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \).

\( \implies \) median \( = 10 + \frac{140 - 49}{133} \times 5 \).
\( = 10 + \frac{91}{133} \times 5 \).
\( = 10 + \frac{13 \times 7}{19 \times 7} \times 5 \).
\( = 10 + \frac{65}{19} \).
\( = 10 + 3.421 \).
\( = 13.421 \).
The median salary is 13.421 thousand rupees.

Question. Find median of the following data:
Class interval | Frequency
130 – 139 | 4
140 – 149 | 9
150 – 159 | 18
160 – 169 | 28
170 – 179 | 24
180 – 189 | 10
190 – 199 | 7
Answer: Sol. Making given distribution continuous by subtracting \( \frac{1}{2} = 0.5 \) from lower limit and adding \( \frac{1}{2} = 0.5 \) to upper limit.
Class interval | \( f \) | \( c.f. \)
129.5 – 139.5 | 4 | 4
139.5 – 149.5 | 9 | 13
149.5 – 159.5 | 18 | 31
159.5 – 169.5 | 28 | 59
169.5 – 179.5 | 24 | 83
179.5 – 189.5 | 10 | 93
189.5 – 199.5 | 7 | 100
\( n = 100 \)

\( \implies \frac{n}{2} = 50 \)
Median class = 159.5 – 169.5,
\( l = 159.5, c.f. = 31, f = 28, h = 10 \)
Median \( = 159.5 + \left( \frac{50 - 31}{28} \right) \times 10 \)
\( = 159.5 + \frac{19}{28} \times 10 = 166.285 \approx 166.3 \)

Question. Find median for the following data:
Wages (in Rs.) | Number of workers
More than 150 | Nil
More than 140 | 12
More than 130 | 27
More than 120 | 60
More than 110 | 105
More than 100 | 124
More than 90 | 141
More than 80 | 150

Answer:
Sol. Wages (in Rs.) | \( f \) | \( c.f. \)
80 – 90 | 9 | 9
90 – 100 | 17 | 26
100 – 110 | 19 | 45
110 – 120 | 45 | 90
120 – 130 | 33 | 123
130 – 140 | 15 | 138
140 – 150 | 12 | 150
\( n = 150 \)
\( \implies \) \( \frac{n}{2} = 75 \)
\( \therefore \) Median class = 110 – 120
\( l = 110, f = 45, c.f. = 45, h = 10 \)
Median \( = l + \frac{\frac{n}{2} - c.f.}{f} \times h \)
\( = 110 + \frac{75 - 45}{45} \times 10 = 116.67 \)

Question. The median of the distribution given below is 14.4. Find the values of \( x \) and \( y \), if the sum of frequency is 20.
Class interval | Frequency
0 – 6 | 4
6 – 12 | \( x \)
12 – 18 | 5
18 – 24 | \( y \)
24 – 30 | 1

Answer:
Sol. Class Interval | \( f \) | \( cf \)
0 – 6 | 4 | 4
6 – 12 | \( x \) | \( 4 + x \)
12 – 18 | 5 | \( 9 + x \)
18 – 24 | \( y \) | \( 9 + x + y \)
24 – 30 | 1 | \( 10 + x + y \)
Total | \( 10 + x + y \)
\( n = 20 \)
\( \implies \) \( \frac{n}{2} = 10 \)
Median = 14.4
\( \therefore \) Median class = 12 – 18
\( \therefore l = 12, c = 4 + x, f = 5, h = 6 \)
Median \( = l + \frac{\frac{n}{2} - c}{f} \times h \)
\( 14.4 = 12 + \frac{10 - (4 + x)}{5} \times 6 \)
\( 2.4 = \frac{6 - x}{5} \times 6 \)
\( \implies \) \( \frac{2.4 \times 5}{6} = 6 - x \)
\( \implies x = 4 \), \( \therefore y = 6 \) (as \( x + y = 10 \))

Question. Weekly income of 600 families is given below. Find the median.
Income in Rs. | Frequency
0 – 1000 | 250
1000 – 2000 | 190
2000 – 3000 | 100
3000 – 4000 | 40
4000 – 5000 | 15
5000 – 6000 | 5

Answer:
Sol. Income in (Rs.) | Number of families (\( f \)) | \( c.f. \)
0 – 1000 | 250 | 250
1000 – 2000 | 190 | 440
2000 – 3000 | 100 | 540
3000 – 4000 | 40 | 580
4000 – 5000 | 15 | 595
5000 – 6000 | 5 | 600
\( n = 600 \)
\( \implies \) \( \frac{n}{2} = 300 \)
\( \therefore \) Median class = 1000 – 2000
\( l = 1000, c = 250, f = 190, h = 1000 \)
Median \( = l + \frac{\frac{n}{2} - c}{f} \times h \)
\( = 1000 + \frac{300 - 250}{190} \times 1000 \)
\( = 1000 + \frac{5000}{19} = 1263.158 \)

Question. If the median of the following frequency distribution is 32.5. Find the values of \( f_1 \) and \( f_2 \).
Classes | Frequency
0 – 10 | \( f_1 \)
10 – 20 | 5
20 – 30 | 9
30 – 40 | 12
40 – 50 | \( f_2 \)
50 – 60 | 3
60 – 70 | 2
Total | 40

Answer:
Sol. Class | Frequency | cumulative frequency
0 – 10 | \( f_1 \) | \( f_1 \)
10 – 20 | 5 | \( 5 + f_1 \)
20 – 30 | 9 | \( 14 + f_1 \)
30 – 40 | 12 | \( 26 + f_1 \)
40 – 50 | \( f_2 \) | \( 26 + f_1 + f_2 \)
50 – 60 | 3 | \( 29 + f_1 + f_2 \)
60 – 70 | 2 | \( 31 + f_1 + f_2 \)
Total \( \sum f_i = 40 \)
Median = 32.5
\( \implies \) median class is 30 – 40.
Now \( 32.5 = 30 + \frac{10}{12} (20 - (14 + f_1)) \)
\( \implies f_1 = 3 \)
Also \( 31 + f_1 + f_2 = 40 \)
\( \implies f_2 = 6 \)

Question. Find mean, median and mode of the following data:
Classes | Frequency
0 – 20 | 6
20 – 40 | 8
40 – 60 | 10
60 – 80 | 12
80 – 100 | 6
100 – 120 | 5
120 – 140 | 3

Answer:
Sol. Classes | Frequency | Cumulative frequency
0 – 20 | 6 | 6
20 – 40 | 8 | 14
40 – 60 | 10 | 24
60 – 80 | 12 | 36
80 – 100 | 6 | 42
100 – 120 | 5 | 47
120 – 140 | 3 | 50
\( n = 50 \)
\( \implies \) \( \frac{n}{2} = 25 \)
Median class = (60 – 80)
\( l = 60, f = 12, c.f. = 24, h = 20 \).
Median \( = l + \frac{\frac{n}{2} - c.f.}{f} \times h \)
\( = 60 + \frac{25 - 24}{12} \times 20 = 61.6 \)
Modal class = (60 – 80) as its frequency is 12
\( h = 20, l = 60, f_1 = 12, f_0 = 10, f_2 = 6 \).
Mode \( = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\( = 60 + \frac{12 - 10}{2 \times 12 - 10 - 6} \times 20 = 65 \)
Now, Mode \( = 3 \text{ Median} - 2 \text{ Mean} \)
\( 65 = 3 (61.6) - 2 \text{ Mean} \)
\( 2 \text{ Mean} = 184.8 - 65 \)
\( 2 \text{ Mean} = 119.8 \)
\( \implies \) Mean \( = \frac{119.8}{2} = 59.9 \)
\( \therefore \) Mean = 59.9; Median = 61.6; Mode = 65

PRACTICE QUESTIONS

Question. The median of set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 2, then the median of the new set
(a) is increased by 2
(b) is decreased by 2
(c) is two times of the original number
(d) Remains the same as that of the original set.
Answer: (d) Remains the same as that of the original set.

Question. The median from the table is
Value: 7, 8, 9, 10, 11, 12, 13
Frequency: 2, 1, 4, 5, 6, 1, 3
(a) 11
(b) 10
(c) 12
(d) 11.5
Answer: (a) 11

Question. The relationship between mean, median and mode for a moderately skewed distribution is
(a) mode = median – 2 mean
(b) mode = 3 median – 2 mean
(c) mode = 2 median – 3 mean
(d) mode = median – mean
Answer: (b) mode = 3 median – 2 mean

Question. Find the medians in an arranged series of an even number of \( 2n \) terms.
Answer: For a series with an even number of terms (\( 2n \)), the median is the arithmetic mean of the \( n \)-th and \( (n + 1) \)-th observations.

Question. For a particular year, the following is the distribution of the age (in yrs.) of primary school teachers in H.P.:
Age (in yrs): 16 – 20, 21 – 25, 26 – 30, 31 – 35, 36 – 40, 41 – 45, 46 – 50
No. of teachers: 11, 32, 51, 49, 27, 6, 4
Find how many teachers are of age less than 31 years.
Answer: Teachers of age less than 31 years are those in the age groups 16 – 20, 21 – 25, and 26 – 30.
Number of teachers \( = 11 + 32 + 51 = 94 \).

Question. Find the median for the following data:
Marks: 5, 7, 9, 10, 12, 14, 16, 17, 19, 20
Frequency: 4, 3, 2, 8, 4, 4, 3, 2, 3, 1

Answer: To find the median, we first find the cumulative frequencies:
Marks (x): 5, 7, 9, 10, 12, 14, 16, 17, 19, 20
Frequency (f): 4, 3, 2, 8, 4, 4, 3, 2, 3, 1
Cumulative Frequency (cf): 4, 7, 9, 17, 21, 25, 28, 30, 33, 34
Here, \( n = 34 \), which is even. The median is the average of the \( \frac{n}{2} \)-th and \( (\frac{n}{2} + 1) \)-th observations.
\( \frac{n}{2} = 17 \)-th observation is 10.
\( \frac{n}{2} + 1 = 18 \)-th observation is 12.
Median \( = \frac{10 + 12}{2} = 11 \).

Question. Calculate the median for the following data:
Classes: 20 – 40, 40 – 60, 60 – 80, 80 – 100, 100 – 120, 120 – 140, 140 – 160
Frequency: 12, 18, 23, 15, 12, 12, 8

Answer: Cumulative frequencies are: 12, 30, 53, 68, 80, 92, 100.
\( n = 100 \), so \( \frac{n}{2} = 50 \).
The median class is 60 – 80 because its cumulative frequency (53) is the first to exceed 50.
\( l = 60, f = 23, cf = 30, h = 20 \).
Median \( = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \)
\( = 60 + \left( \frac{50 - 30}{23} \right) \times 20 \)
\( = 60 + \frac{400}{23} \approx 60 + 17.39 = 77.39 \).

Question. The median of the following data is 525. Find the values of \( x \) and \( y \), if total frequency is 100:
C.I.: 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
[CBSE 2020]

Answer: Total frequency \( = 100 \).
\( \implies 2 + 5 + x + 12 + 17 + 20 + y + 9 + 7 + 4 = 100 \)
\( \implies 76 + x + y = 100 \implies x + y = 24 \).
Median \( = 525 \), so median class is 500 – 600.
\( l = 500, f = 20, h = 100, cf = 36 + x \).
Median \( = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h \)
\( 525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100 \)
\( 25 = (14 - x) \times 5 \)
\( 5 = 14 - x \implies x = 9 \).
From \( x + y = 24 \), \( y = 24 - 9 = 15 \).
Thus, \( x = 9 \) and \( y = 15 \).

Question. Compute the median from the following data:
Mid-value: 115, 125, 135, 145, 155, 165, 175, 185, 195
Frequency: 6, 25, 48, 72, 116, 60, 38, 22, 3

Answer: The class size \( h = 10 \). The class intervals are: 110–120, 120–130, 130–140, 140–150, 150–160, 160–170, 170–180, 180–190, 190–200.
Cumulative Frequencies: 6, 31, 79, 151, 267, 327, 365, 387, 390.
\( n = 390, \frac{n}{2} = 195 \).
Median class is 150 – 160.
\( l = 150, f = 116, cf = 151, h = 10 \).
Median \( = 150 + \left( \frac{195 - 151}{116} \right) \times 10 \)
\( = 150 + \frac{440}{116} \approx 150 + 3.79 = 153.79 \).

INTEGRATED (MIXED) QUESTIONS

Question. The mean age of combined group of men and women is 30 years. If the mean of the age of men and women are respectively 32 and 27, then the percentage of women in the group is
(a) 30
(b) 20
(c) 50
(d) 40
Answer: (d) 40

Question. The middle most observation of a statistical data has value which is called
(a) mean of the data
(b) mode of the data
(c) median of the data
(d) None of the options
Answer: (c) median of the data

Question. The mean and median of the same data are 24 and 26 respectively. The value of mode is:
(a) 23
(b) 26
(c) 25
(d) 30
Answer: (d) 30

Question. Which of the following is not a measure of central tendency:
(a) Mean
(b) Median
(c) Class interval
(d) Mode
Answer: (c) Class interval

Question. The mean of first \( n \) natural numbers is
(a) \( \frac{n}{2} \)
(b) \( \frac{n + 1}{2} \)
(c) \( \frac{n + 1}{n} \)
(d) \( \frac{n(n + 1)}{2} \)
Answer: (b) \( \frac{n + 1}{2} \)

Question. The mean of 25 observations is 9. If each observation is increased by 4, then the new mean is
(a) 14
(b) 17
(c) 15
(d) 13
Answer: (d) 13

Question. The mean of 5 numbers is 18. One number is excluded their mean becomes 16. Then the excluded number is
(a) 23
(b) 24
(c) 25
(d) 26
Answer: (d) 26

Question. The mean and median of same data are 24 and 26 respectively. Find mode of same data.
(a) 30
(b) 31
(c) 40
(d) 41
Answer: (a) 30

Question. Mode and mean of a data are \( 12k \) and \( 15k \). Find median of the data.
(a) \( 7k \)
(b) \( 17k \)
(c) \( 14k \)
(d) \( 13k \)
Answer: (c) \( 14k \)

Question. Write the relationship between three measures of central tendency-Mean, Median and Mode.
(a) 2 medians = mode + 3 mean
(b) 3 medians = mode + 2 mean
(c) medians = 2 means – mode
(d) 3 medians = 2 mode + mean
Answer: (b) 3 medians = mode + 2 mean