Statistics: 2 marks Questions (CBSE Class 10)

15 different 2 marks questions on Statistics from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (69)2 marks (15)3 marks (27)4 marks (14)5 marks (73)

Q42 marksVery Short AnswerStatistics CBSE 2022 · Basic 430/1/1

Find the mode of the following frequency distribution :
Class: 0–20, 20–40, 40–60, 60–80, 80–100
Frequency: 8, 7, 12, 5, 3

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Answer:

\frac{145}{3} \approx 48.33

Highest frequency is 12, so the modal class is 40–60
$l = 40$ , $h = 20$ , $f_{1} = 12$ , $f_{0} = 7$ , $f_{2} = 5$
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 40 + \frac{12 - 7}{24 - 7 - 5} \times 20$
$= 40 + \frac{5}{12} \times 20 = 40 + 8.33 = 48.33$

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q52 marksVery Short AnswerStatistics CBSE 2022 · Basic 430/2/1

Find the mode of the following frequency distribution :
Class: 20–30, 30–40, 40–50, 50–60, 60–70
Frequency: 25, 30, 45, 42, 35

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Answer: Mode ≈ 48.33

Maximum frequency 45, so modal class is 40–50.
$l = 40$ , $h = 10$ , $f_{1} = 45$ , $f_{0} = 30$ , $f_{2} = 42$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 40 + \frac{15}{90 - 30 - 42} \times 10$
$= 40 + \frac{150}{18} = 40 + 8.33 = 48.33$ .

Also asked in: 2022 Basic 430/2/2, 2022 Basic 430/2/3

Q32 marksVery Short AnswerStatistics CBSE 2022 · Basic 430/3/1

Find mode of the following frequency distribution :
Class: 100–110, 110–120, 120–130, 130–140, 140–150
Frequency: 5, 9, 8, 11, 7

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Answer: Mode

= 130 + \frac{30}{7} \approx 134.29

Highest frequency is 11, so the modal class is 130–140.
$l = 130$ , $h = 10$ , $f_{1} = 11$ , $f_{0} = 8$ , $f_{2} = 7$
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 130 + \frac{11 - 8}{22 - 8 - 7} \times 10$
$= 130 + \frac{30}{7} = 134 \frac{2}{7} \approx 134.29$

Also asked in: 2022 Basic 430/3/2, 2022 Basic 430/3/3

Q32 marksVery Short AnswerStatistics CBSE 2022 · Basic 430/4/1

Using the empirical relationship between the three measures of central tendency, find the median of a distribution, whose mean is 169 and mode is 175.

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Answer: Median = 171

Empirical relation: $3 Median = Mode + 2 Mean$ .
$3 Median = 175 + 2 \times 169 = 175 + 338 = 513$ .
Median $= \frac{513}{3} = 171$ .

Also asked in: 2022 Basic 430/4/2

Q42 marksVery Short AnswerStatistics CBSE 2022 · Basic 430/4/1

The frequency distribution table of agriculture holding in a village is given below :
Area of Land (in hectares): 1 – 3, 3 – 5, 5 – 7, 7 – 9, 9 – 11, 11 – 13
Number of families: 20, 45, 80, 55, 40, 12
Find the modal agriculture holding per family.

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Answer: Modal holding

= \frac{37}{6} \approx 6.17

hectares

Highest frequency is 80, so the modal class is 5 – 7.
$l = 5$ , $f_{1} = 80$ , $f_{0} = 45$ , $f_{2} = 55$ , $h = 2$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 5 + \frac{35}{160 - 45 - 55} \times 2$ .
$= 5 + \frac{35}{60} \times 2 = 5 + \frac{7}{6} = \frac{37}{6} \approx 6.17$ hectares.

Also asked in: 2022 Basic 430/4/3

Q32 marksVery Short AnswerStatistics CBSE 2022 · Basic 430/4/2

Find the modal and median classes of the following distribution.
Class : 0 – 20, 20 – 40, 40 – 60, 60 – 80, 80 – 100
Frequency : 11, 22, 19, 13, 7

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Answer: Modal class: 20 – 40; median class: 40 – 60

Highest frequency is 22, so the modal class is 20 – 40.
Cumulative frequencies: 11, 33, 52, 65, 72; $N = 72$ , $\frac{N}{2} = 36$ .
The first cumulative frequency greater than 36 is 52, so the median class is 40 – 60.

Q32 marksVery Short AnswerStatistics CBSE 2022 · Basic 430/4/3

Find the sum of the lower limit of the modal class and upper limit of the median class for the following distribution.
Class : 50 – 55, 55 – 60, 60 – 65, 65 – 70, 70 – 75, 75 – 80
Frequency : 10, 15, 8, 13, 9, 5

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Answer: 120

Highest frequency is 15, so the modal class is 55 – 60; lower limit = 55.
Cumulative frequencies: 10, 25, 33, 46, 55, 60; $N = 60$ , $\frac{N}{2} = 30$ .
Median class is 60 – 65; upper limit = 65.
Sum $= 55 + 65 = 120$ .

Q42 marksVery Short AnswerStatistics CBSE 2022 · Standard 30/1/1

Find the mode of the following frequency distribution :
Class: 10–20, 20–30, 30–40, 40–50, 50–60
Frequency: 15, 10, 12, 17, 4

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Answer:

\frac{385}{9} \approx 42.78

Highest frequency is 17, so the modal class is 40–50.
$l = 40$ , $h = 10$ , $f_{1} = 17$ , $f_{0} = 12$ , $f_{2} = 4$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 40 + \frac{17 - 12}{34 - 12 - 4} \times 10$
$= 40 + \frac{50}{18} = 40 + 2.78 = 42.78$ (approx.)

Also asked in: 2022 Standard 30/1/2, 2022 Standard 30/1/3

Q52 marksVery Short AnswerStatistics CBSE 2022 · Standard 30/2/1

If mode of the following frequency distribution is 55, then find the value of $x$ .
Class: 0 – 15, 15 – 30, 30 – 45, 45 – 60, 60 – 75, 75 – 90
Frequency: 10, 7, $x$ , 15, 10, 12

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Answer:

x = 5

Mode 55 lies in 45 – 60, so this is the modal class.
$l = 45$ , $h = 15$ , $f_{1} = 15$ , $f_{0} = x$ , $f_{2} = 10$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$
$55 = 45 + \frac{15 - x}{30 - x - 10} \times 15$
$10 (20 - x) = 15 (15 - x)$
$200 - 10 x = 225 - 15 x \Rightarrow 5 x = 25 \Rightarrow x = 5$ .

Also asked in: 2022 Standard 30/2/2, 2022 Standard 30/2/3

Q32 marksVery Short AnswerStatistics CBSE 2022 · Standard 30/3/1

Find the mode of the given frequency distribution :
Class: 15 – 25, 25 – 35, 35 – 45, 45 – 55, 55 – 65, 65 – 75
Frequency: 6, 11, 22, 23, 14, 5

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Answer: 46

Highest frequency is 23, so the modal class is 45 – 55.
$l = 45$ , $h = 10$ , $f_{1} = 23$ , $f_{0} = 22$ , $f_{2} = 14$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 45 + \frac{23 - 22}{46 - 22 - 14} \times 10$
$= 45 + \frac{1}{10} \times 10 = 46$ .

Also asked in: 2022 Standard 30/3/2, 2022 Standard 30/3/3

Q12 marksVery Short AnswerStatistics CBSE 2022 · Standard 30/4/1

The mode of a grouped frequency distribution is 75 and the modal class is 65-80. The frequency of the class preceding the modal class is 6 and the frequency of the class succeeding the modal class is 8. Find the frequency of the modal class.

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Answer: 10

Let the frequency of the modal class be $f_{1}$ . Here $l = 65$ , $h = 15$ , $f_{0} = 6$ , $f_{2} = 8$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$
$75 = 65 + \frac{f _{1} - 6}{2 f _{1} - 14} \times 15$
$10 (2 f_{1} - 14) = 15 (f_{1} - 6)$
$20 f_{1} - 140 = 15 f_{1} - 90 \Rightarrow 5 f_{1} = 50$
Frequency of the modal class $= 10$ .

Also asked in: 2022 Standard 30/4/3

Q42 marksVery Short AnswerStatistics CBSE 2022 · Standard 30/4/1

Find the mean of the following frequency distribution :
Class: 1–5, 5–9, 9–13, 13–17
Frequency: 4, 8, 7, 6

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Answer: 9.4

Class marks $x_{i}$ : 3, 7, 11, 15; frequencies $f_{i}$ : 4, 8, 7, 6.
$\sum f_{i} = 25$
$\sum f_{i} x_{i} = 12 + 56 + 77 + 90 = 235$
Mean $= \frac{235}{25} = 9.4$

Q52 marksVery Short AnswerStatistics CBSE 2022 · Standard 30/4/2

For the following frequency distribution, find the mode :
Class: 25–30, 30–35, 35–40, 40–45, 45–50
Frequency: 12, 5, 14, 8, 9

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Answer: 38

Highest frequency is 14, so the modal class is 35–40.
$l = 35$ , $h = 5$ , $f_{1} = 14$ , $f_{0} = 5$ , $f_{2} = 8$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 35 + \frac{9}{28 - 13} \times 5$
$= 35 + \frac{9}{15} \times 5 = 35 + 3 = 38$

Q62 marksVery Short AnswerStatistics CBSE 2022 · Standard 30/4/2

If the mean of the following frequency distribution is 18, then find the missing frequency ‘f’.
Class: 11–13, 13–15, 15–17, 17–19, 19–21, 21–23, 23–25
Frequency: 3, 6, 9, 13, f, 5, 4

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Answer:

f = 8

Class marks: 12, 14, 16, 18, 20, 22, 24.
$\sum f_{i} = 40 + f$
$\sum f_{i} x_{i} = 36 + 84 + 144 + 234 + 20 f + 110 + 96 = 704 + 20 f$
$\frac{704 + 20 f}{40 + f} = 18$
$704 + 20 f = 720 + 18 f \Rightarrow 2 f = 16$
$f = 8$

Q52 marksVery Short AnswerStatistics CBSE 2022 · Standard 30/4/3

Find the median of the following distribution :
Marks: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60
Number of students: 5, 8, 20, 15, 7, 5

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Answer: 28.5

Cumulative frequencies: 5, 13, 33, 48, 55, 60; $N = 60$ , $\frac{N}{2} = 30$ .
Median class is 20–30: $l = 20$ , $c f = 13$ , $f = 20$ , $h = 10$ .
Median $= 20 + \frac{30 - 13}{20} \times 10 = 20 + 8.5 = 28.5$

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