Statistics: 3 marks Questions (CBSE Class 10)

27 different 3 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)

Q303 marksShort AnswerStatistics CBSE 2024 · Standard 30/2/1

In a test, the marks obtained by 100 students (out of 50) are given below :
Marks obtained : 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50
Number of students : 12, 23, 34, 25, 6
Find the mean marks of the students.

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Answer: 24

Class marks $x_{i}$ : 5, 15, 25, 35, 45
$f_{i} x_{i}$ : 60, 345, 850, 875, 270
$\sum f_{i} = 100$ , $\sum f_{i} x_{i} = 2400$
Mean $= \frac{2400}{100} = 24$ marks

Also asked in: 2024 Standard 30/2/2, 2024 Standard 30/2/3

Q283 marksShort AnswerStatistics CBSE 2023 · Basic 430/5/1

Find mean of the following data :
Class: 0 – 15, 15 – 30, 30 – 45, 45 – 60, 60 – 75, 75 – 90
Frequency: 12, 15, 11, 20, 16, 6

Show answer & solution

Answer: Mean = 43.3125 (about 43.31)

Class marks $x_{i}$ : 7.5, 22.5, 37.5, 52.5, 67.5, 82.5.
Take $a = 52.5$ , $h = 15$ , $u_{i} = \frac{x _{i} - 52.5}{15}$ : $- 3, - 2, - 1, 0, 1, 2$ .
$f_{i} u_{i}$ : $- 36, - 30, - 11, 0, 16, 12$ ; $Σ f_{i} u_{i} = - 49$ , $Σ f_{i} = 80$ .
Mean $= 52.5 + 15 \times \frac{- 49}{80} = 52.5 - 9.1875 = 43.3125$ .

Also asked in: 2023 Basic 430/5/2, 2023 Basic 430/5/3

Q273 marksShort AnswerStatistics CBSE 2023 · Basic 430/6/1

A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household :
Family size: 1-3, 3-5, 5-7, 7-9, 9-11
Number of Families: 7, 8, 2, 2, 1
Find the median of this data.

Show answer & solution

Answer: 3.75

Cumulative frequencies: 7, 15, 17, 19, 20. $N = 20$ , $\frac{N}{2} = 10$ .
Median class = 3-5 (first cf $\geq 10$ ).
$l = 3$ , $c f = 7$ , $f = 8$ , $h = 2$ .
Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 3 + \frac{10 - 7}{8} \times 2 = 3 + 0.75 = 3.75$ .

Also asked in: 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q313 marksShort AnswerStatistics CBSE 2023 · Standard 30/5/1

Find the mean of the following frequency distribution :
Classes: 25–30, 30–35, 35–40, 40–45, 45–50, 50–55, 55–60
Frequency: 14, 22, 16, 6, 5, 3, 4

Show answer & solution

Answer: Mean ≈ 36.86

Class marks $x_{i}$ : 27.5, 32.5, 37.5, 42.5, 47.5, 52.5, 57.5; $\sum f_{i} = 70$ .
Take a = 42.5, h = 5, $u_{i} = \frac{x _{i} - 42.5}{5}$ : $- 3, - 2, - 1, 0, 1, 2, 3$ .
$f_{i} u_{i}$ : $- 42, - 44, - 16, 0, 5, 6, 12$ ; $\sum f_{i} u_{i} = - 79$ .
Mean $= 42.5 + 5 \times \frac{- 79}{70} = 42.5 - 5.64 = 36.86$ (approx.).

Also asked in: 2023 Standard 30/5/3

Q263 marksShort AnswerStatistics CBSE 2023 · Standard 30/5/2

Find the mean of the following distribution :
Classes: 0–15, 15–30, 30–45, 45–60, 60–75, 75–90
Frequency: 17, 20, 18, 21, 15, 9

Show answer & solution

Answer: Mean = 41.1

Class marks $x_{i}$ : 7.5, 22.5, 37.5, 52.5, 67.5, 82.5; $\sum f_{i} = 100$ .
$f_{i} x_{i}$ : 127.5, 450, 675, 1102.5, 1012.5, 742.5; $\sum f_{i} x_{i} = 4110$ .
Mean $= \frac{4110}{100} = 41.1$ .

Q73 marksShort AnswerStatistics CBSE 2022 · Basic 430/1/1

The frequency distribution given below shows the weight of 40 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
Number of Students: 9, 5, 8, 9, 6, 3

Show answer & solution

Answer: 53.75 kg

Cumulative frequencies: 9, 14, 22, 31, 37, 40
$N = 40$ , $\frac{N}{2} = 20$ , so the median class is 50–55
$l = 50$ , $c f = 14$ , $f = 8$ , $h = 5$
Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 50 + \frac{20 - 14}{8} \times 5$
$= 50 + 3.75 = 53.75$ kg

Q103 marksShort AnswerStatistics CBSE 2022 · Basic 430/1/1

The following table shows the age of patients admitted in a hospital during a particular week :
Age (in years): 5–15, 15–25, 25–35, 35–45, 45–55, 55–65
Number of Patients: 5, 12, 20, 24, 15, 4
Find the mean age of the patients.

Show answer & solution

Answer: 35.5 years

Class marks $x_{i}$ : 10, 20, 30, 40, 50, 60
$f_{i} x_{i}$ : 50, 240, 600, 960, 750, 240
$\sum f_{i} = 80$ , $\sum f_{i} x_{i} = 2840$
Mean $= \frac{2840}{80} = 35.5$ years

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

Q73 marksShort AnswerStatistics CBSE 2022 · Basic 430/1/2

Determine the median marks for the following data :
Marks: 0–10, 10–20, 20–30, 30–40, 40–50
Number of Students: 3, 8, 15, 10, 8

Show answer & solution

Answer:

\frac{82}{3} \approx 27.33

Cumulative frequencies: 3, 11, 26, 36, 44
$N = 44$ , $\frac{N}{2} = 22$ , so the median class is 20–30
$l = 20$ , $c f = 11$ , $f = 15$ , $h = 10$
Median $= 20 + \frac{22 - 11}{15} \times 10 = 20 + \frac{22}{3}$
$= 27.33$ (approx.)

Q103 marksShort AnswerStatistics CBSE 2022 · Basic 430/1/3

Weekly income of 110 families is given below :
Weekly Income (in ₹): 5000–6000, 6000–7000, 7000–8000, 8000–9000, 9000–10000
Number of Families: 18, 30, 28, 19, 15
Find the median weekly income for this data.

Show answer & solution

Answer: ₹7250

Cumulative frequencies: 18, 48, 76, 95, 110
$N = 110$ , $\frac{N}{2} = 55$ , so the median class is 7000–8000
$l = 7000$ , $c f = 48$ , $f = 28$ , $h = 1000$
Median $= 7000 + \frac{55 - 48}{28} \times 1000 = 7000 + 250$
Median = ₹7250

Q93 marksShort AnswerStatistics CBSE 2022 · Basic 430/2/1

Find the mean of the following frequency distribution :
Class: 10–15, 15–20, 20–25, 25–30, 30–35
Frequency: 4, 10, 5, 6, 5

Show answer & solution

Answer: Mean ≈ 22.17

Class marks $x_{i}$ : 12.5, 17.5, 22.5, 27.5, 32.5.
$f_{i} x_{i}$ : 50, 175, 112.5, 165, 162.5.
$\sum f_{i} = 30$ , $\sum f_{i} x_{i} = 665$ .
Mean $= \frac{665}{30} = 22.17$ (approx.).

Also asked in: 2022 Basic 430/2/2

Q103 marksShort AnswerStatistics CBSE 2022 · Basic 430/2/1

The median of following frequency distribution is 25. Find the value of $x$ .
Class: 0–10, 10–20, 20–30, 30–40, 40–50
Frequency: 6, 9, 10, 8, $x$

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Answer:

x = 7

$N = 6 + 9 + 10 + 8 + x = 33 + x$ .
Median 25 lies in class 20–30: $l = 20$ , $c f = 15$ , $f = 10$ , $h = 10$ .
$25 = 20 + \frac{\frac{33 + x}{2} - 15}{10} \times 10$
$5 = \frac{33 + x}{2} - 15 \Rightarrow \frac{33 + x}{2} = 20 \Rightarrow x = 7$ .
Check: N = 40, N/2 = 20 lies in 20–30.

Also asked in: 2022 Basic 430/2/3

Q103 marksShort AnswerStatistics CBSE 2022 · Basic 430/2/2

The median of the following frequency distribution is 35. Find the value of $x$ .
Class: 0–10, 10–20, 20–30, 30–40, 40–50
Frequency: 6, 3, $x$ , 12, 19

Show answer & solution

Answer:

x = 10

$N = 6 + 3 + x + 12 + 19 = 40 + x$ .
Median 35 lies in class 30–40: $l = 30$ , $c f = 9 + x$ , $f = 12$ , $h = 10$ .
$35 = 30 + \frac{\frac{40 + x}{2} - ( 9 + x )}{12} \times 10$
$6 = 20 + \frac{x}{2} - 9 - x = 11 - \frac{x}{2} \Rightarrow x = 10$ .
Check: N = 50, N/2 = 25; cf before 30–40 is 19, so 30–40 is the median class.

Q83 marksShort AnswerStatistics CBSE 2022 · Basic 430/2/3

Find mean of the following frequency distribution :
Class: 0–20, 20–40, 40–60, 60–80, 80–100
Frequency: 6, 8, 5, 9, 7

Show answer & solution

Answer: Mean

= \frac{362}{7} \approx 51.71

Class marks $x_{i}$ : 10, 30, 50, 70, 90.
$f_{i} x_{i}$ : 60, 240, 250, 630, 630.
$\sum f_{i} = 35$ , $\sum f_{i} x_{i} = 1810$ .
Mean $= \frac{1810}{35} = \frac{362}{7} \approx 51.71$ .

Q93 marksShort AnswerStatistics CBSE 2022 · Basic 430/3/1

The mileage (km/l) of 50 cars was recorded by a dealer and tabulated as given below :
Mileage (in km/l): 10–12, 12–14, 14–16, 16–18, 18–20
Number of Cars: 13, 18, 10, 7, 2
Find mean of the above distribution.

Show answer & solution

Answer: Mean mileage

= 13.68

km/l

Class marks $x_{i}$ : 11, 13, 15, 17, 19; $f_{i}$ : 13, 18, 10, 7, 2; $\sum f_{i} = 50$
$f_{i} x_{i}$ : 143, 234, 150, 119, 38; $\sum f_{i} x_{i} = 684$
Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{684}{50} = 13.68$ km/l

Also asked in: 2022 Basic 430/3/2

Q103 marksShort AnswerStatistics CBSE 2022 · Basic 430/3/1

Determine median of the following frequency distribution :
Class: 15–20, 20–25, 25–30, 30–35, 35–40, 40–45
Frequency: 8, 13, 21, 12, 5, 4

Show answer & solution

Answer: Median

= 27.5

Cumulative frequencies: 8, 21, 42, 54, 59, 63; $n = 63$ , $\frac{n}{2} = 31.5$
Median class is 25–30: $l = 25$ , $c f = 21$ , $f = 21$ , $h = 5$
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 25 + \frac{31.5 - 21}{21} \times 5 = 25 + 2.5 = 27.5$

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

Q103 marksShort AnswerStatistics CBSE 2022 · Basic 430/3/3

Determine the mean of the following frequency distribution :
Class: 100–110, 110–120, 120–130, 130–140, 140–150, 150–160
Frequency: 5, 4, 6, 8, 3, 4

Show answer & solution

Answer: Mean

= 129

Class marks $x_{i}$ : 105, 115, 125, 135, 145, 155; $\sum f_{i} = 30$
$f_{i} x_{i}$ : 525, 460, 750, 1080, 435, 620; $\sum f_{i} x_{i} = 3870$
Mean $= \frac{3870}{30} = 129$

Q73 marksShort AnswerStatistics CBSE 2022 · Standard 30/1/1

For what value of x, is the median of the following frequency distribution 34.5 ?
Class: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60, 60–70
Frequency: 3, 5, 11, 10, x, 3, 2

Show answer & solution

Answer:

x = 13

Cumulative frequencies: 3, 8, 19, 29, $29 + x$ , $32 + x$ , $34 + x$ ; so $N = 34 + x$ .
Median 34.5 lies in the class 30–40: $l = 30$ , $c f = 19$ , $f = 10$ , $h = 10$ .
$34.5 = 30 + \frac{\frac{N}{2} - 19}{10} \times 10 = 11 + \frac{N}{2}$
$\frac{N}{2} = 23.5 \Rightarrow N = 47$
$34 + x = 47 \Rightarrow x = 13$

Also asked in: 2022 Standard 30/1/3

Q103 marksShort AnswerStatistics CBSE 2022 · Standard 30/1/1

Following is the daily expenditure on lunch by 30 employees of a company :
Daily Expenditure (in Rupees): 100–120, 120–140, 140–160, 160–180, 180–200
Number of Employees: 8, 3, 8, 6, 5
Find the mean daily expenditure of the employees.

Show answer & solution

Answer: ₹148

Class marks $x_{i}$ : 110, 130, 150, 170, 190; $Σ f_{i} = 30$ .
$f_{i} x_{i}$ : 880, 390, 1200, 1020, 950; $Σ f_{i} x_{i} = 4440$ .
Mean $= \frac{Σ f _{i} x _{i}}{Σ f _{i}} = \frac{4440}{30} = 148$
Mean daily expenditure is ₹148.

Also asked in: 2022 Standard 30/1/2

Q93 marksShort AnswerStatistics CBSE 2022 · Standard 30/1/2

If the median of the following data is 33 then, find the value of m :
Class: 0–15, 15–30, 30–45, 45–60, 60–75, 75–90
Frequency: 17, 35, 40, 18, m, 2

Show answer & solution

Answer:

m = 8

Cumulative frequencies: 17, 52, 92, 110, $110 + m$ , $112 + m$ ; so $N = 112 + m$ .
Median 33 lies in the class 30–45: $l = 30$ , $c f = 52$ , $f = 40$ , $h = 15$ .
$33 = 30 + \frac{\frac{112 + m}{2} - 52}{40} \times 15$
$3 = \frac{15}{40} (4 + \frac{m}{2}) \Rightarrow 8 = 4 + \frac{m}{2}$
$m = 8$ (check: $N = 120$ , $\frac{N}{2} = 60$ lies in 52–92, so 30–45 is the median class).

Q103 marksShort AnswerStatistics CBSE 2022 · Standard 30/1/3

The following table gives the production yield of wheat of farms of a village :
Production Yield (in kg/ha): 50–60, 60–70, 70–80, 80–90, 90–100
Number of Farms: 7, 12, 11, 8, 2
Find the mean production yield, using assumed mean method.

Show answer & solution

Answer: 71.5 kg/ha

Class marks $x_{i}$ : 55, 65, 75, 85, 95. Take assumed mean $a = 75$ ; $d_{i} = x_{i} - 75$ : $- 20, - 10, 0, 10, 20$ .
$f_{i} d_{i}$ : $- 140, - 120, 0, 80, 40$ ; $Σ f_{i} d_{i} = - 140$ , $Σ f_{i} = 40$ .
Mean $= a + \frac{Σ f _{i} d _{i}}{Σ f _{i}} = 75 + \frac{- 140}{40} = 75 - 3.5 = 71.5$
Mean production yield is 71.5 kg/ha.

Q93 marksShort AnswerStatistics CBSE 2022 · Standard 30/2/1

The mean of the following frequency distribution is 25. Find the value of f.
Class: 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50
Frequency: 5, 18, 15, f, 6

Show answer & solution

Answer: f = 16

Class marks $x_{i}$ : 5, 15, 25, 35, 45.
$f_{i} x_{i}$ : 25, 270, 375, 35f, 270.
$\sum f_{i} = 44 + f$ , $\sum f_{i} x_{i} = 940 + 35 f$ .
Mean $= \frac{940 + 35 f}{44 + f} = 25$
$940 + 35 f = 1100 + 25 f \Rightarrow 10 f = 160 \Rightarrow f = 16$ .

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

Q9 (OR) (OR)3 marksShort AnswerStatistics CBSE 2022 · Standard 30/2/1

Find the mean of the following data using assumed mean method :
Class: 0 – 5, 5 – 10, 10 – 15, 15 – 20, 20 – 25
Frequency: 8, 7, 10, 13, 12

Show answer & solution

Answer: Mean = 13.9

Class marks $x_{i}$ : 2.5, 7.5, 12.5, 17.5, 22.5. Take assumed mean $a = 12.5$ .
$d_{i} = x_{i} - a$ : $- 10, - 5, 0, 5, 10$ .
$f_{i} d_{i}$ : $- 80, - 35, 0, 65, 120$ .
$\sum f_{i} = 50$ , $\sum f_{i} d_{i} = 70$ .
Mean $= a + \frac{\sum f _{i} d _{i}}{\sum f _{i}} = 12.5 + \frac{70}{50} = 12.5 + 1.4 = 13.9$ .

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

Q103 marksShort AnswerStatistics CBSE 2022 · Standard 30/2/1

Heights of 50 students of class X of a school are recorded and following data is obtained :
Height (in cm): 130-135, 135-140, 140-145, 145-150, 150-155, 155-160
Number of Students: 4, 11, 12, 7, 10, 6
Find the median height of the students.

Show answer & solution

Answer: Median height

= 144 \frac{1}{6}

\approx 144.17

Cumulative frequencies: 4, 15, 27, 34, 44, 50.
$n = 50$ , $\frac{n}{2} = 25$ , so the median class is 140-145.
$l = 140$ , $c f = 15$ , $f = 12$ , $h = 5$ .
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 140 + \frac{25 - 15}{12} \times 5$
$= 140 + \frac{50}{12} = 140 + 4.17 = 144.17$ cm (approx.).

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

Q93 marksShort AnswerStatistics CBSE 2022 · Standard 30/3/1

The weights (in kg) of 50 wild animals of a National Park were recorded and the following data was obtained :
Weight (in kg): 100 – 110, 110 – 120, 120 – 130, 130 – 140, 140 – 150
Number of animals: 4, 12, 23, 8, 3
Find the mean weight (in kg) of animals, using assumed mean method.

Show answer & solution

Answer: 123.8 kg

Class marks $x_{i}$ : 105, 115, 125, 135, 145. Take assumed mean $a = 125$ .
$d_{i} = x_{i} - 125$ : $- 20, - 10, 0, 10, 20$ .
$f_{i} d_{i}$ : $- 80, - 120, 0, 80, 60$ ; $\sum f_{i} d_{i} = - 60$ , $\sum f_{i} = 50$ .
Mean $= a + \frac{\sum f _{i} d _{i}}{\sum f _{i}} = 125 + \frac{- 60}{50} = 125 - 1.2 = 123.8$ kg.

Also asked in: 2022 Standard 30/3/2

Q103 marksShort AnswerStatistics CBSE 2022 · Standard 30/3/1

For the following frequency distribution, find the median :
Class: 1400 – 1550, 1550 – 1700, 1700 – 1850, 1850 – 2000
Frequency: 6, 13, 25, 10

Show answer & solution

Answer: 1748

Cumulative frequencies: 6, 19, 44, 54. $N = 54$ , $\frac{N}{2} = 27$ .
Median class is 1700 – 1850: $l = 1700$ , $c f = 19$ , $f = 25$ , $h = 150$ .
Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 1700 + \frac{27 - 19}{25} \times 150 = 1700 + 48 = 1748$ .

Also asked in: 2022 Standard 30/3/3

Q83 marksShort AnswerStatistics CBSE 2022 · Standard 30/3/2

The percentage of marks obtained by 100 students in an examination are given below :
Percentage of Marks: 30 – 35, 35 – 40, 40 – 45, 45 – 50, 50 – 55, 55 – 60, 60 – 65
Number of Students: 16, 14, 18, 20, 18, 12, 2
Determine the median percentage of marks.

Show answer & solution

Answer: 45.5

Cumulative frequencies: 16, 30, 48, 68, 86, 98, 100. $N = 100$ , $\frac{N}{2} = 50$ .
Median class is 45 – 50: $l = 45$ , $c f = 48$ , $f = 20$ , $h = 5$ .
Median $= 45 + \frac{50 - 48}{20} \times 5 = 45 + 0.5 = 45.5$ .

Q103 marksShort AnswerStatistics CBSE 2022 · Standard 30/3/3

Find the mean of the following frequency distribution :
Class: 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60
Frequency: 12, 18, 27, 20, 17, 6

Show answer & solution

Answer: 28

Class marks $x_{i}$ : 5, 15, 25, 35, 45, 55; $\sum f_{i} = 100$ .
$f_{i} x_{i}$ : 60, 270, 675, 700, 765, 330; $\sum f_{i} x_{i} = 2800$ .
Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{2800}{100} = 28$ .

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