CBSE Class 10 Mathematics Statistics VBQs Set F

OBJECTIVE TYPE QUESTIONS

Question. 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 the deviations 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)
Explanation: In the given formula, \( a \) is assumed mean from class marks \( (x_i) \) and \( d_i = x_i - a \). Therefore, \( d_i \) is the deviation of class mark (mid-value) from the assumed mean ‘\( a \)’.

Question. While computing mean of grouped data, we assume that the frequencies are:
(a) evenly distributed over all the classes
(b) centred at the class marks of the classes
(c) centred at the upper limits of the classes
(d) centred at the lower limits of the classes
Answer: (b)
Explanation: In grouping the data from ungrouped data, all the observations between lower and upper limits of class marks are taken in one group then mid-value or class mark is taken for further calculation. Therefore, frequencies or observations must be centred at the class marks of the classes.

Question. If \( x_i \)’s are the mid-points of the class intervals of grouped data, \( f_i \)’s are the corresponding frequencies and \( \bar{x} \) is the mean, then \( \sum (f_i (x_i - \bar{x})) \) is equal to:
(a) 0
(b) −1
(c) 1
(d) 2
Answer: (a)
Explanation: \( \because \bar{x} = \frac{\sum_{i=1}^{n} f_i x_i}{n} \)
\( \therefore \sum_{i=1}^{n} f_i x_i = n \bar{x} \) ...(i)
\( \sum_{i=1}^{n} \bar{x} = \bar{x} + \bar{x} + \bar{x} + ... n \text{ times} \)
\( \Rightarrow \sum_{i=1}^{n} \bar{x} = n \bar{x} \) ...(ii)
From equations (i) and (ii), we have
\( \Rightarrow \sum_{i=1}^{n} f_i x_i = \sum_{i=1}^{n} \bar{x} \)
\( \Rightarrow \sum_{i=1}^{n} (f_i x_i - \bar{x}) = 0 \)

Question. For the following distribution:
Class: 0 – 5, 5 – 10, 10 – 15, 15 – 20, 20 – 25
Frequency: 10, 15, 12, 20, 9
the sum of lower limits of median class and modal class is:
(a) 15
(b) 25
(c) 30
(d) 35
Answer: (b)
Explanation:
Class | Frequency | Cumulative frequency
0 – 5 | 10 | 10
5 – 10 | 15 | 25
10 – 15 | 12 | 37
15 – 20 | 20 | 57
20 – 25 | 9 | 66
The modal class is the class having the maximum frequency. The maximum frequency 20 belongs to class (15–20). Here, \( \sum f_i = n = 66 \). So, \( \frac{n}{2} = \frac{66}{2} = 33 \). 33 lies in the class 10–15. Therefore, 10–15 is the median class. So, sum of lower limits of (15–20) and (10–15) is (15 + 10) = 25.

Question. Consider the following frequency distribution:
Class: 0–5, 6–11, 12–17, 18–23, 24–29
Frequency: 13, 10, 15, 8, 11
the upper limit of the median class is:
(a) 7
(b) 17.5
(c) 18
(d) 18.5
Answer: (b)
Explanation:
Class | Frequency | Cumulative frequency
0.5 – 5.5 | 13 | 13
5.5 – 11.5 | 10 | 23
11.5 – 17.5 | 15 | 38
17.5 – 23.5 | 8 | 46
23.5 – 29.5 | 11 | 57
The median of 57 (odd) observations = \( \frac{57 + 1}{2} = \frac{58}{2} = 29^{th} \) term. \( 29^{th} \) term lies in class 11.5 – 17.5. So, upper limit is 17.5.

Question. For the following distribution:
Marks | Number of students
Below 10 | 3
Below 20 | 12
Below 30 | 27
Below 40 | 57
Below 50 | 75
Below 60 | 80
the modal class is:
(a) 10 – 20
(b) 20 – 30
(c) 30 – 40
(d) 50 – 60
Answer: (c)
Explanation:
Marks | Number of students | \( f_i \)
0 – 10 | 3 – 0 = 3 | 3
10 – 20 | 12 – 3 = 9 | 9
20 – 30 | 27 – 12 = 15 | 15
30 – 40 | 57 – 27 = 30 | 30
40 – 50 | 75 – 57 = 18 | 18
50 – 60 | 80 – 75 = 5 | 5
Modal class has maximum frequency (30) in class 30 – 40.

Case-based MCQs 

Read the following text and answer the questions that follow, on the basis of the same.
COVID-19 Pandemic
The COVID-19 pandemic, also known as coronavirus pandemic, is an ongoing pandemic of corona virus disease caused by the transmission of severe acute respiratory syndrome corona virus 2 (SARS-CoV-2) among humans. The following tables shows the age distribution of case admitted during a day in two different hospitals
Table 1
Age (in years): 5 – 15, 15 – 25, 25 – 35, 35 – 45, 45 – 55, 55 – 65
No. of cases: 6, 11, 21, 23, 14, 5
Table 2
Age (in years): 5 – 15, 15 – 25, 25 – 35, 35 – 45, 45 – 55, 55 – 65
No. of cases: 8, 16, 10, 42, 24, 12

Question. Refer to table 1: The average age for which maximum cases occurred is:
(a) 32.24
(b) 34.36
(c) 36.82
(d) 42.24
Answer: (c)
Explanation: Since, highest frequency is 23. So, modal class is 35 – 45.
Now, \( Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
Here, \( l = 35, h = 10, f_1 = 23, f_0 = 21, f_2 = 14 \),
\( \Rightarrow Mode = 35 + \left( \frac{23 - 21}{46 - 21 - 14} \right) \times 10 = 35 + \frac{2}{11} \times 10 = 35 + \frac{20}{11} = 35 + 1.81 = 36.818 \approx 36.82 \)

Question. Refer to table 1: The upper limit of modal class is:
(a) 15
(b) 25
(c) 35
(d) 45
Answer: (d)

Question. Refer to table 1: The mean of the given data is:
(a) 26.2
(b) 32.4
(c) 33.5
(d) 35.4
Answer: (d)
Explanation:
Age (in years) | Class marks (\( x_i \)) | Frequency (\( f_i \)) | Deviation \( d_i = (x_i - a) \) | \( f_i d_i \)
5 – 15 | 10 | 6 | –20 | –120
15 – 25 | 20 → \( a \) | 11 | –10 | –110
25 – 35 | 30 | 21 | 0 | 0
35 – 45 | 40 | 23 | 10 | 230
45 – 55 | 50 | 14 | 20 | 280
55 – 65 | 60 | 5 | 30 | 150
Total | | \( \sum f_i = 80 \) | | \( \sum f_i d_i = 430 \)
Now, \( Mean (\bar{x}) = a + \frac{\sum f_i d_i}{\sum f_i} = 30 + \frac{430}{80} = 30 + 5.375 = 35.375 = 35.4 \)

Question. Refer to table 2: The mode of the given data is:
(a) 41.4
(b) 48.2
(c) 55.3
(d) 64.6
Answer: (a)
Explanation: Here, Modal class is 35 - 45. (As highest frequency is 42)
Now, \( Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
Here, \( l = 35, f_1 = 42, f_0 = 10, f_2 = 24, h = 10 \)
\( Mode = 35 + \frac{42 - 10}{2 \times 42 - 10 - 24} \times 10 = 35 + \frac{32}{50} \times 10 = 35 + 6.4 = 41.4 \)

Question. Refer to table 2: The median of the given data is:
(a) 32.7
(b) 40.2
(c) 42.3
(d) 48.6
Answer: (b)
Explanation:
Age (in years) | Frequency (\( f_i \)) | Cumulative frequency (c.f.)
5 – 15 | 8 | 8
15 – 25 | 16 | 24
25 – 35 | 10 | 34
35 – 45 | 42 | 76
45 – 55 | 24 | 100
55 – 65 | 12 | 112
Total | \( n = 112 \) |
Now, \( \frac{n}{2} = \frac{112}{2} = 56 \).
\( l = 35 \) (lower limit of median class), \( c.f. = 34 \) (Preceding), \( f = 42 \), \( h = 10 \)
\( Median = l + \left( \frac{\frac{n}{2} - c.f.}{f} \right) \times h = 35 + \left( \frac{56 - 34}{42} \right) \times 10 = 35 + \left( \frac{22}{42} \right) \times 10 = 35 + 5.24 = 40.24 \approx 40.2 \)

Read the following text and answer the following question on the basis of the same.
Electricity Energy Consumption
Electricity energy consumption is the form of energy consumption that uses electric energy. Global electricity consumption continues to increase faster than world population, leading to an increase in the average amount of electricity consumed per person.
A survey is conducted for 56 families of colony A and 80 families of colony B. The following tables gives the weekly consumption of electricity of these families.
Colony A: Table 2
Weekly consumption (in units): 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60
No. of families: 16, 12, 18, 6, 4, 0
Colony B: Table 2
Weekly consumption (in units): 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60
No. of families: 1, 5, 10, 20, 40, 5

Question. Refer to data received from Colony A: The median weekly consumption is:
(a) 12 units
(b) 16 units
(c) 20 units
(d) None of these
Answer: (c)
Explanation:
Weekly consumption | Frequency (\( f_i \)) | Cumulative frequency (c.f.)
0 – 10 | 16 | 16
10 – 20 | 12 | 28
20 – 30 | 18 | 46
30 – 40 | 6 | 52
40 – 50 | 4 | 56
50 – 60 | 0 | 56
\( n = 56, \frac{n}{2} = 28 \). Median class is 10–20. \( l = 10, c.f. = 16, f = 12, h = 10 \).
\( Median = 10 + \left( \frac{28 - 16}{12} \right) \times 10 = 10 + \left( \frac{12}{12} \right) \times 10 = 10 + 10 = 20 \) units.

Question. Refer to data received from Colony A: The mean weekly consumption is:
(a) 19.64 units
(b) 22.5 units
(c) 26 units
(d) None of these
Answer: (a)
Explanation:
Weekly consumption | \( x_i \) | \( f_i \) | \( f_i x_i \)
0-10 | 5 | 16 | 80
10-20 | 15 | 12 | 180
20-30 | 25 | 18 | 450
30-40 | 35 | 6 | 210
40-50 | 45 | 4 | 180
50-60 | 55 | 0 | 0
Total | | \( \sum f_i = 56 \) | \( \sum f_i x_i = 1100 \)
\( Mean = \frac{1100}{56} = 19.64 \)

Question. Refer to data received from Colony A: The modal class of the above data is:
(a) 0 – 10
(b) 10 – 20
(c) 20 – 30
(d) 30 – 40
Answer: (c)
Explanation: Modal class is the class with highest frequency i.e., 20 – 30

Question. Refer to data received from Colony B (Table 2): The modal weekly consumption is:
(a) 38.2 units
(b) 43.6 units
(c) 26 units
(d) 32 units
Answer: (b)
Explanation: Here, modal class is 40 - 50 as highest frequency is 40.
\( l = 40, f_1 = 40, f_0 = 20, f_2 = 5, h = 10 \)
\( Mode = 40 + \left( \frac{40 - 20}{2 \times 40 - 20 - 5} \right) \times 10 = 40 + \frac{200}{55} = 40 + 3.63 = 43.63 \approx 43.6 \) units

Question. The mean weekly consumption is:
(a) 15.65 units
(b) 32.8 units
(c) 38.75 units
(d) 48 units
Answer: (c)
Explanation:
Weekly consumption (in units) | \(x_i\) | \(f_i\) | \(f_i x_i\)
0-10 | 5 | 1 | 0
10-20 | 15 | 5 | 75
20-30 | 25 | 10 | 250
30-40 | 35 | 20 | 700
40-50 | 45 | 40 | 1800
50-60 | 55 | 5 | 275
Total | | \(\sum f_i = 80\) | \(\sum f_i x_i = 3105\)
Mean = \(\frac{\sum f_i x_i}{\sum f_i}\) = \(\frac{3105}{80}\) = 38.81 ≈ 38.8 units

III. Read the following text and answer the questions that follow, on the basis of the same.

The maximum bowling speeds, in km per hour of 33 players at a cricket coaching centre are given as follows.
Speed (in km/h): 85 – 100, 100 – 115, 115 – 130, 130 – 145
Number of players: 11, 9, 8, 5

Question. What is the modal class of the given data?
(a) 85 – 100
(b) 100 – 115
(c) 115 – 130
(d) 130 – 145
Answer: (a)
Explanation: Modal class is the class with highest frequency i.e., 85 – 100

Question. What is the value of class interval for the given data set?
(a) 10
(b) 15
(c) 5
(d) 20
Answer: (b)
Explanation: The value of class interval = 100 – 85 = 15
again 115 – 100 = 15
and 130 – 115 = 15
145 – 130 = 15

Question. What is the median class of the given data?
(a) 85 – 100
(b) 100 – 115
(c) 115 – 130
(d) 130 – 145
Answer: (b)
Explanation: n = Number of observations = 33
Median of 33 observations = 16.5 observation, which lies in class 100 – 115.

Question. What is the median of bowling speed?
(a) 109.17 km/hr (Approx.)
(b) 109.71 km/hr (Approx.)
(c) 107.17 km/hr (Approx.)
(d) 109.19 km/hr (Approx.)
Answer: (a)
Explanation: Median = \(l + \left( \frac{\frac{n}{2} - c.f.}{f} \right) \times h\)
\(l = 100, f = 9, c.f. = 11, h = 100 - 85 = 15\)
Median = \(100 + \left( \frac{\frac{33}{2} - 11}{9} \right) \times 15\)
= \(100 + \left( \frac{16.5 - 11}{9} \right) \times 15\)
= \(100 + \frac{5.5 \times 15}{9}\)
= \(100 + \frac{82.5}{9}\)
= 100 + 9.166 = 109.17 km/h (Approx.)
Hence, the median bowling speed is 109.17 km/h (Approx.)

Question. What is the sum of lower limit of modal class and upper limit of median class?
(a) 100
(b) 200
(c) 300
(d) 400
Answer: (b)
Explanation: Lower limit of modal class = 85
and upper limit of median class = 115
∴ sum = 85 + 115 = 200

IV. Read the following text and answer the questions that follows, on the basis of the same:

100 m RACE
A stopwatch was used to find the time that it took a group of students to run 100 m.
Time (in sec) | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100
No. of students | 8 | 10 | 13 | 6 | 3

Question. Estimate the mean time taken by a student to finish the race.
(a) 54
(b) 63
(c) 43
(d) 50
Answer: (c)
Explanation:
Time (in sec) | x | f | fx
0-20 | 10 | 8 | 80
20-40 | 30 | 10 | 300
40-60 | 50 | 13 | 650
60-80 | 70 | 6 | 420
80-100 | 90 | 3 | 270
Total | | 40 | 1720
Mean = \(\frac{1720}{40}\) = 43

Question. What will be the upper limit of the modal class ?
(a) 20
(b) 40
(c) 60
(d) 80
Answer: (c)
Explanation: Modal class = 40 – 60
Upper limit = 60

Question. The construction of cumulative frequency table is useful in determining the:
(a) Mean
(b) Median
(c) Mode
(d) All of the above
Answer: (b)
Explanation: The construction of c.f. table is useful in determining the median.

Question. The sum of lower limits of median class and modal class is:
(a) 60
(b) 100
(c) 80
(d) 140
Answer: (c)
Explanation: Median class = 40 – 60
Modal class = 40 – 60
Therefore, the sum of the lower limits of median and modal class = 40 + 40 = 80

Question. How many students finished the race within 1 minute?
(a) 18
(b) 37
(c) 31
(d) 8
Answer: (c)
Explanation: Number of students who finished the race within 1 minute = 8 + 10 + 13 = 31

Very Short Answer Type Questions 

Question. Find the class-marks of the classes 10 – 25 and 35 – 55.
Answer: Class mark of 10 – 25 = \(\frac{10 + 25}{2}\) = \(\frac{35}{2}\) = 17.5
and Class mark of 35 – 55 = \(\frac{35 + 55}{2}\) = \(\frac{90}{2}\) = 45

Question. Find the class marks of the classes 20 – 50 and 35 – 60.
Answer: Class mark of 20 – 50 = \(\frac{20 + 50}{2}\) = \(\frac{70}{2}\) = 35
and Class mark of 35 – 60 = \(\frac{35 + 60}{2}\) = \(\frac{95}{2}\) = 47.5

Question. Consider the following frequency distribution of the heights of 60 students of a class.
Heights (in cm): 150 – 155, 155 – 160, 160 – 165, 165 – 170, 170 – 175, 175 – 180
No. of students: 15, 13, 10, 8, 9, 5
Find the upper limit of the median class in the given data.
Answer:
Heights (in cm) | No. of students | Cumulative frequency
150 – 155 | 15 | 15
155 – 160 | 13 | 15 + 13 = 28
160 – 165 | 10 | 28 + 10 = 38
165 – 170 | 8 | 38 + 8 = 46
170 – 175 | 9 | 46 + 9 = 55
175 – 180 | 5 | 55 + 5 = 60
Since total frequency is 60. \(\frac{n}{2}\) = 30. And cumulative frequency greater than or equal to 30 lies in class 160 - 165. So, median class is 160 - 165. ∴ Upper limit of median class is 165.

Question. Following distribution gives cumulative frequencies of 'more than type' :
Marks obtained: More than or equal to 5, More than or equal to 10, More than or equal to 15, More than or equal to 20
Number of students (cumulative frequency): 30, 23, 8, 2
Change the above data to a continuous grouped frequency distribution.
Answer:
C.I. | 5 – 10 | 10–15 | 15–20 | More than 20
f | 7 | 15 | 6 | 2

Question. In the following frequency distribution, find the median class.
Height (in cm): 140 – 145, 145 – 150, 150 – 155, 155 – 160, 160 – 165, 165 – 170
Frequency: 5, 15, 25, 30, 15, 10
Answer:
Height | Frequency | c.f.
140 – 145 | 5 | 5
145 – 150 | 15 | 20
150 – 155 | 25 | 45
155 – 160 | 30 | 75
160 – 165 | 15 | 90
165 – 170 | 10 | 100
N = \(\sum f\) = 100 ⇒ \(\frac{N}{2}\) = \(\frac{100}{2}\) = 50. The cumulative frequency just greater than 50 is 75 and the corresponding class is 155 – 160. Hence, median class is 155 – 160.

Short Answer Type Questions

Question. Find the mean of the following distribution :
Class: 3 – 5, 5 – 7, 7 – 9, 9 – 11, 11 – 13
Frequency: 5, 10, 10, 7, 8
Answer:
Class | Frequency (f) | Mid-Value (x) | f × x
3 – 5 | 5 | 4 | 20
5 – 7 | 10 | 6 | 60
7 – 9 | 10 | 8 | 80
9 – 11 | 7 | 10 | 70
11 – 13 | 8 | 12 | 96
Total | \(\sum f = 40\) | | \(\sum fx = 326\)
Mean = \(\frac{\sum fx}{\sum f}\) = \(\frac{326}{40}\) = 8.15.

Question. Find the mode of the following data:
Class: 0 – 20, 20 – 40, 40 – 60, 60 – 80, 80 – 100, 100 – 120, 120 – 140
Frequency: 6, 8, 10, 12, 6, 5, 3
Answer: Since the modal class is the class having the maximum frequency. ∴ Modal class = 60 – 80. ∴ Mode = \(l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\)
Hence, \(l = 60, f_1 = 12, f_0 = 10, f_2 = 6\) and \(h = 20\)
Mode = \(60 + \frac{12 - 10}{2 \times 12 - 10 - 6} \times 20\) = \(60 + \frac{2 \times 20}{24 - 16}\) = \(60 + \frac{40}{8}\) = 60 + 5 = 65.

Question. Compute the mode for the following frequency distribution :
Size of items (in cm): 0 – 4, 4 – 8, 8 – 12, 12 – 16, 16 – 20, 20 – 24, 24 – 28
Frequency: 5, 7, 9, 17, 12, 10, 6
Answer: Here, Modal class = 12 – 16. ∴ \(l = 12, f_1 = 17, f_0 = 9, f_2 = 12\) and \(h = 4\)
Mode = \(l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\)
= \(12 + \left( \frac{17 - 9}{2 \times 17 - 9 - 12} \right) \times 4\)
= \(12 + \frac{8 \times 4}{13}\) = 12 + 2.46 = 14.46 (Approx.)

Question. Find the mode of the following frequency distribution :
Class: 15 – 20, 20 – 25, 25 – 30, 30 – 35, 35 – 40, 40 – 45
Frequency: 3, 8, 9, 10, 3, 2
Answer: Here, modal class = 30 – 35. ∴ \(l = 30, f_0 = 9, f_1 = 10, f_2 = 3\) and \(h = 5\)
Mode = \(l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\)
= \(30 + \left( \frac{10 - 9}{2 \times 10 - 9 - 3} \right) \times 5\)
= \(30 + \frac{5}{8}\) = 30 + 0.625 = 30.625.

Question. Find the mode of the following distribution :
Class: 25 – 30, 30 – 35, 35 – 40, 40 – 45, 45 – 50, 50 – 55
Frequency: 25, 34, 50, 42, 38, 14
Answer: Maximum frequency = 50, class (modal) = 35 – 40. Mode = \(l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\) = \(35 + \frac{50 - 34}{100 - 34 - 42} \times 5\) = \(35 + \frac{16}{24} \times 5\) = 38.33 (Approx.)

Question. Find the unknown values in the following table:
Class Interval | Frequency | Cumulative Frequency
0 – 10 | 5 | 5
10 – 20 | 7 | \(x_1\)
20 – 30 | \(x_2\) | 18
30 – 40 | 5 | \(x_3\)
40 – 50 | \(x_4\) | 30
Answer: \(x_1\) = 5 + 7 = 12
\(x_2\) = 18 – \(x_1\) = 18 – 12 = 6
\(x_3\) = 18 + 5 = 23
and \(x_4\) = 30 – \(x_3\) = 30 – 23 = 7