Statistics: 5 marks Questions (CBSE Class 10)

73 different 5 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)

Q355 marksLong AnswerStatistics CBSE 2026 · Basic 430/5/1

Find mean and mode of the following data :

Class	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	5	4	10	13	12	10	6

Show answer & solution

Answer: Mean

\approx 47.83

; Mode = 47.5

Class marks: 15, 25, 35, 45, 55, 65, 75; N = 60.
Take a = 45, h = 10, $u = \frac{x - 45}{10}$ : −3, −2, −1, 0, 1, 2, 3.
$f u$ : −15, −8, −10, 0, 12, 20, 18; $\sum f u = 17$
Mean $= 45 + \frac{17}{60} \times 10 = 45 + 2.83 = 47.83$ (approx.)
Modal class 40-50: $l = 40$ , $f_{1} = 13$ , $f_{0} = 10$ , $f_{2} = 12$ , $h = 10$
Mode $= 40 + \frac{13 - 10}{2 \times 13 - 10 - 12} \times 10 = 40 + \frac{3}{4} \times 10 = 47.5$

Q335 marksLong AnswerStatistics CBSE 2026 · Basic 430/5/2

Find mean and mode of the following data :

Class	0-15	15-30	30-45	45-60	60-75	75-90	90-105
Frequency	4	6	8	10	12	7	3

Show answer & solution

Answer: Mean = 53.4; Mode

= 60 + \frac{30}{7} \approx 64.29

Class marks: 7.5, 22.5, 37.5, 52.5, 67.5, 82.5, 97.5; N = 50.
Take a = 52.5, h = 15, $u = \frac{x - 52.5}{15}$ : −3, −2, −1, 0, 1, 2, 3.
$f u$ : −12, −12, −8, 0, 12, 14, 9; $\sum f u = 3$
Mean $= 52.5 + \frac{3}{50} \times 15 = 52.5 + 0.9 = 53.4$
Modal class 60-75: $l = 60$ , $f_{1} = 12$ , $f_{0} = 10$ , $f_{2} = 7$ , $h = 15$
Mode $= 60 + \frac{12 - 10}{24 - 10 - 7} \times 15 = 60 + \frac{30}{7} \approx 64.29$

Q325 marksLong AnswerStatistics CBSE 2026 · Basic 430/5/3

Find the missing frequencies p and q in the following frequency distribution, when sum of frequencies is 40 and mean is 19 :

Class	0-5	5-10	10-15	15-20	20-25	25-30	30-35
Frequency	2	5	6	p	10	q	4

Show answer & solution

Answer: p = 7, q = 6

Sum of frequencies: 2 + 5 + 6 + p + 10 + q + 4 = 40, so p + q = 13.
Class marks: 2.5, 7.5, 12.5, 17.5, 22.5, 27.5, 32.5.
$\sum f x = 5 + 37.5 + 75 + 17.5 p + 225 + 27.5 q + 130 = 472.5 + 17.5 p + 27.5 q$
Mean $= \frac{\sum f x}{40} = 19 \Rightarrow 472.5 + 17.5 p + 27.5 q = 760 \Rightarrow 7 p + 11 q = 115$
Using $p = 13 - q$ : $91 + 4 q = 115 \Rightarrow q = 6$ , $p = 7$

Q355 marksLong AnswerStatistics CBSE 2026 · Standard 30/1/1

The mean of the following frequency distribution is 35. Find the values of $x$ and $y$ , if the sum of frequencies is 25 :

Class	Frequency
0-10	1
10-20	$x$
20-30	5
30-40	7
40-50	$y$
50-60	3
60-70	1

Show answer & solution

Answer:

x = 3

y = 5

Sum of frequencies: $1 + x + 5 + 7 + y + 3 + 1 = 25$ , so $x + y = 8$ .
Class marks: 5, 15, 25, 35, 45, 55, 65.
$\sum f_{i} x_{i} = 5 + 15 x + 125 + 245 + 45 y + 165 + 65 = 605 + 15 x + 45 y$ .
Mean $= \frac{605 + 15 x + 45 y}{25} = 35$ , so $15 x + 45 y = 270$ , i.e. $x + 3 y = 18$ .
Subtracting $x + y = 8$ : $2 y = 10$ , $y = 5$ ; then $x = 3$ .

Q345 marksLong AnswerStatistics CBSE 2026 · Standard 30/1/2

The monthly expenditure on fruits in 200 families of a Housing Society is given below. Find the value of $x$ and also find the mode and mean expenditure on fruits.

Monthly Expenditure (in ₹)	No. of Families
1000-1500	24
1500-2000	40
2000-2500	33
2500-3000	28
3000-3500	$x$
3500-4000	22
4000-4500	16
4500-5000	7

Show answer & solution

Answer:

x = 30

; mode ≈ ₹ 1847.83; mean = ₹ 2662.50

$24 + 40 + 33 + 28 + x + 22 + 16 + 7 = 200$ , so $170 + x = 200$ and $x = 30$ .
Mode: modal class 1500-2000 (highest frequency 40). $l = 1500$ , $f_{1} = 40$ , $f_{0} = 24$ , $f_{2} = 33$ , $h = 500$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 1500 + \frac{16}{23} \times 500 = 1500 + 347.83 = 1847.83$ (approx.).
Mean: class marks 1250, 1750, 2250, 2750, 3250, 3750, 4250, 4750.
$\sum f_{i} x_{i} = 30000 + 70000 + 74250 + 77000 + 97500 + 82500 + 68000 + 33250 = 532500$ .
Mean $= \frac{532500}{200} = 2662.5$ .
Mode ≈ ₹ 1847.83 and mean = ₹ 2662.50.

Q335 marksLong AnswerStatistics CBSE 2026 · Standard 30/1/3

The marks obtained by 80 students of class X in a mock test of Mathematics are given below in the table. Find median and the mode of the data :

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

Show answer & solution

Answer: Median = 52; Mode = 55

Class frequencies (by subtraction): 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 (total 80).
Cumulative frequencies: 3, 8, 15, 25, 37, 52, 64, 70, 72, 80.
Median: $\frac{N}{2} = 40$ , so the median class is 50-60 with $l = 50$ , $c f = 37$ , $f = 15$ , $h = 10$ .
Median $= 50 + \frac{40 - 37}{15} \times 10 = 50 + 2 = 52$ .
Mode: modal class 50-60 with $f_{1} = 15$ , $f_{0} = 12$ , $f_{2} = 12$ .
Mode $= 50 + \frac{15 - 12}{30 - 12 - 12} \times 10 = 50 + \frac{3}{6} \times 10 = 55$ .

Q355 marksLong AnswerStatistics CBSE 2026 · Standard 30/2/1

An SBI health insurance agent found the following data for distribution of ages of 100 policy holders. The health insurance policies are given to persons of age 15 years and onwards, but less than 60 years.

Age (in yrs)	Number of policy holders
15 - 20	2
20 - 25	4
25 - 30	18
30 - 35	21
35 - 40	33
40 - 45	11
45 - 50	3
50 - 55	6
55 - 60	2

Find the modal age and median age of the policy holders.

Show answer & solution

Answer: Modal age ≈ 36.76 years; median age ≈ 35.76 years

Mode: modal class 35 - 40 (highest frequency 33); l = 35, $f_{1} = 33$ , $f_{0} = 21$ , $f_{2} = 11$ , h = 5.
Mode $= 35 + \frac{33 - 21}{2 ( 33 ) - 21 - 11} \times 5 = 35 + \frac{12}{34} \times 5 = 35 + 1.76 = 36.76$ years
Cumulative frequencies: 2, 6, 24, 45, 78, 89, 92, 98, 100; N = 100, $\frac{N}{2} = 50$ .
Median class 35 - 40; l = 35, cf = 45, f = 33, h = 5.
Median $= 35 + \frac{50 - 45}{33} \times 5 = 35 + 0.76 = 35.76$ years

Also asked in: 2026 Standard 30/2/3

Q325 marksLong AnswerStatistics CBSE 2026 · Standard 30/2/2

If the median of the following distribution is 32.5, then find the values of x and y.

Class	Frequency
0 - 10	x
10 - 20	5
20 - 30	9
30 - 40	12
40 - 50	y
50 - 60	3
60 - 70	2
Total	40

Show answer & solution

Answer: x = 3, y = 6

x + 5 + 9 + 12 + y + 3 + 2 = 40, so x + y = 9.
N = 40, $\frac{N}{2} = 20$ . Median 32.5 lies in class 30 - 40.
l = 30, cf = x + 14, f = 12, h = 10.
$32.5 = 30 + \frac{20 - ( x + 14 )}{12} \times 10$
$2.5 = \frac{( 6 - x ) \times 10}{12}$ , so 6 - x = 3 and x = 3.
y = 9 - 3 = 6

Q325 marksLong AnswerStatistics CBSE 2026 · Standard 30/3/1

The median of the following data is 137. Find the values of x and y, given that total of frequencies is 68.
Class: 65 – 85, 85 – 105, 105 – 125, 125 – 145, 145 – 165, 165 – 185, 185 – 205
Frequency: 4, 5, x, 20, 14, y, 4

Show answer & solution

Answer:

x = 13

y = 8

Total: $4 + 5 + x + 20 + 14 + y + 4 = 68$ , so $x + y = 21$
Cumulative frequencies: 4, 9, $9 + x$ , $29 + x$ , $43 + x$ , ...
$\frac{n}{2} = 34$ ; median 137 lies in 125 – 145: $l = 125$ , $c f = 9 + x$ , $f = 20$ , $h = 20$
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h$ : $137 = 125 + \frac{34 - 9 - x}{20} \times 20$
$12 = 25 - x$ , so $x = 13$
$y = 21 - 13 = 8$

Also asked in: 2026 Standard 30/3/2

Q32 (OR) (OR)5 marksLong AnswerStatistics CBSE 2026 · Standard 30/3/1

Find mean and mode of the following distribution :
Class: 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60, 60 – 70
Frequency: 3, 6, 11, 10, 13, 3, 4

Show answer & solution

Answer: Mean

= 34.8

; Mode

= \frac{550}{13} \approx 42.31

Class marks $x_{i}$ : 5, 15, 25, 35, 45, 55, 65
$f_{i} x_{i}$ : 15, 90, 275, 350, 585, 165, 260
$\sum f_{i} = 50$ , $\sum f_{i} x_{i} = 1740$
Mean $= \frac{1740}{50} = 34.8$
Modal class 40 – 50 (highest frequency 13): $l = 40$ , $f_{1} = 13$ , $f_{0} = 10$ , $f_{2} = 3$ , $h = 10$
Mode $= 40 + \frac{13 - 10}{2 \times 13 - 10 - 3} \times 10 = 40 + \frac{30}{13} \approx 42.31$

Also asked in: 2026 Standard 30/3/2

Q335 marksLong AnswerStatistics CBSE 2026 · Standard 30/3/3

Find the mean and the mode of the following frequency distribution :
Class: 0 – 15, 15 – 30, 30 – 45, 45 – 60, 60 – 75, 75 – 90, 90 – 105
Frequency: 9, 15, 35, 20, 11, 13, 17

Show answer & solution

Answer: Mean

= 52

; Mode

= \frac{270}{7} \approx 38.57

Class marks $x_{i}$ : 7.5, 22.5, 37.5, 52.5, 67.5, 82.5, 97.5
$f_{i} x_{i}$ : 67.5, 337.5, 1312.5, 1050, 742.5, 1072.5, 1657.5
$\sum f_{i} = 120$ , $\sum f_{i} x_{i} = 6240$
Mean $= \frac{6240}{120} = 52$
Modal class 30 – 45 (highest frequency 35): $l = 30$ , $f_{1} = 35$ , $f_{0} = 15$ , $f_{2} = 20$ , $h = 15$
Mode $= 30 + \frac{35 - 15}{2 \times 35 - 15 - 20} \times 15 = 30 + \frac{20}{35} \times 15 = 30 + \frac{60}{7} \approx 38.57$

Q355 marksLong AnswerStatistics CBSE 2026 · Standard 30/4/1

The median of the following data is 50 and sum of all frequencies is 90 :

Class :	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80	80 – 90
Frequency :	p	15	25	20	q	8	10

Find the values of p and q.

Show answer & solution

Answer: p = 5, q = 7

$p + 15 + 25 + 20 + q + 8 + 10 = 90 \Rightarrow p + q = 12$
Median 50 lies in class 50 – 60: l = 50, f = 20, h = 10, cf = p + 40, N/2 = 45
$50 = 50 + \frac{45 - ( p + 40 )}{20} \times 10$
$\Rightarrow 5 - p = 0 \Rightarrow p = 5$
q = 12 $-$ 5 = 7

Q35 (OR) (OR)5 marksLong AnswerStatistics CBSE 2026 · Standard 30/4/1

Find mean and mode of the following distribution :

Class :	0 – 15	15 – 30	30 – 45	45 – 60	60 – 75	75 – 90	90 – 105
Frequency :	4	8	11	14	10	7	6

Show answer & solution

Answer: Mean = 53.25; Mode =

\frac{360}{7} \approx 51.43

Class marks: 7.5, 22.5, 37.5, 52.5, 67.5, 82.5, 97.5
$Σ f = 60$
$Σ f x = 30 + 180 + 412.5 + 735 + 675 + 577.5 + 585 = 3195$
Mean $= \frac{3195}{60} = 53.25$
Modal class 45 – 60: l = 45, $f_{1} = 14$ , $f_{0} = 11$ , $f_{2} = 10$ , h = 15
Mode $= 45 + \frac{14 - 11}{28 - 11 - 10} \times 15 = 45 + \frac{45}{7} = \frac{360}{7} \approx 51.43$

Q335 marksLong AnswerStatistics CBSE 2026 · Standard 30/4/2

Find mean and mode of the following frequency distribution :

Class :	5 – 15	15 – 25	25 – 35	35 – 45	45 – 55	55 – 65
Frequency :	11	20	25	22	12	10

Show answer & solution

Answer: Mean = 33.4; Mode = 31.25

Class marks: 10, 20, 30, 40, 50, 60; $Σ f = 100$
$Σ f x = 110 + 400 + 750 + 880 + 600 + 600 = 3340$
Mean $= \frac{3340}{100} = 33.4$
Modal class 25 – 35: l = 25, $f_{1} = 25$ , $f_{0} = 20$ , $f_{2} = 22$ , h = 10
Mode $= 25 + \frac{25 - 20}{50 - 20 - 22} \times 10 = 25 + 6.25 = 31.25$

Q33 (OR) (OR)5 marksLong AnswerStatistics CBSE 2026 · Standard 30/4/2

The median of the following data is 32.5, find the missing frequencies $x$ and y :

Class :	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70	Total
Frequency :	$x$	5	9	12	y	3	2	40

Show answer & solution

Answer:

x = 3

, y = 6

$x + 5 + 9 + 12 + y + 3 + 2 = 40 \Rightarrow x + y = 9$
Median 32.5 lies in 30 – 40: l = 30, f = 12, h = 10, cf $= x + 14$ , N/2 = 20
$32.5 = 30 + \frac{20 - ( x + 14 )}{12} \times 10$
$2.5 = \frac{( 6 - x ) \times 10}{12} \Rightarrow 6 - x = 3 \Rightarrow x = 3$
y = 9 $-$ 3 = 6

Q345 marksLong AnswerStatistics CBSE 2026 · Standard 30/4/3

Find mean and mode of the following frequency distribution :

Class :	10 – 30	30 – 50	50 – 70	70 – 90	90 – 110	110 – 130	130 – 150
Frequency :	6	8	12	10	14	11	9

Show answer & solution

Answer: Mean =

\frac{594}{7} \approx 84.86

; Mode =

\frac{710}{7} \approx 101.43

Class marks: 20, 40, 60, 80, 100, 120, 140; $Σ f = 70$
$Σ f x = 120 + 320 + 720 + 800 + 1400 + 1320 + 1260 = 5940$
Mean $= \frac{5940}{70} = \frac{594}{7} \approx 84.86$
Modal class 90 – 110: l = 90, $f_{1} = 14$ , $f_{0} = 10$ , $f_{2} = 11$ , h = 20
Mode $= 90 + \frac{14 - 10}{28 - 10 - 11} \times 20 = 90 + \frac{80}{7} = \frac{710}{7} \approx 101.43$

Q34 (OR) (OR)5 marksLong AnswerStatistics CBSE 2026 · Standard 30/4/3

If the median of the distribution given below is 28.5, find the values of $x$ and y.

Class :	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	Total
Frequency :	5	$x$	20	15	y	5	60

Show answer & solution

Answer:

x = 8

, y = 7

$5 + x + 20 + 15 + y + 5 = 60 \Rightarrow x + y = 15$
Median 28.5 lies in 20 – 30: l = 20, f = 20, h = 10, cf $= 5 + x$ , N/2 = 30
$28.5 = 20 + \frac{30 - ( 5 + x )}{20} \times 10 = 20 + \frac{25 - x}{2}$
$17 = 25 - x \Rightarrow x = 8$
y = 15 $-$ 8 = 7

Q345 marksLong AnswerStatistics CBSE 2026 · Standard 30/5/1

The mean of the following frequency distribution is 28. If sum of all frequencies is 100, then find the values of p and q :

Class Interval	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
Frequency	12	p	27	20	q	6

Show answer & solution

Answer: p = 18, q = 17

Sum of frequencies: $12 + p + 27 + 20 + q + 6 = 100$ , so $p + q = 35$ ... (1)
Class marks: 5, 15, 25, 35, 45, 55
$\sum f_{i} x_{i} = 60 + 15 p + 675 + 700 + 45 q + 330 = 1765 + 15 p + 45 q$
Mean $= \frac{1765 + 15 p + 45 q}{100} = 28$ , so $15 p + 45 q = 1035$ , i.e. $p + 3 q = 69$ ... (2)
(2) − (1): $2 q = 34$ , so $q = 17$ and $p = 18$ .

Q34 (OR) (OR)5 marksLong AnswerStatistics CBSE 2026 · Standard 30/5/1

Find median and mode of the following distribution :

Class Interval	0 – 15	15 – 30	30 – 45	45 – 60	60 – 75	75 – 90	90 – 105
Frequency	15	10	12	9	8	10	6

Show answer & solution

Answer: Median = 42.5; mode = 11.25

N = 15 + 10 + 12 + 9 + 8 + 10 + 6 = 70, $\frac{N}{2} = 35$
Cumulative frequencies: 15, 25, 37, 46, 54, 64, 70
Median class 30 – 45: l = 30, cf = 25, f = 12, h = 15
Median $= 30 + \frac{35 - 25}{12} \times 15 = 30 + 12.5 = 42.5$
Modal class 0 – 15 (highest frequency 15): l = 0, $f_{1} = 15$ , $f_{0} = 0$ , $f_{2} = 10$ , h = 15
Mode $= 0 + \frac{15 - 0}{2 ( 15 ) - 0 - 10} \times 15 = \frac{15}{20} \times 15 = 11.25$

Q355 marksLong AnswerStatistics CBSE 2026 · Standard 30/5/2

Find the mean and mode of the following frequency distribution :

Class Interval :	400-450	450-500	500-550	550-600	600-650	650-700
Frequency :	15	18	20	23	22	12

Show answer & solution

Answer: Mean = 550; mode = 587.5

N = 15 + 18 + 20 + 23 + 22 + 12 = 110
Class marks: 425, 475, 525, 575, 625, 675. Take a = 575, h = 50; $u_{i} = - 3, - 2, - 1, 0, 1, 2$
$\sum f_{i} u_{i} = - 45 - 36 - 20 + 0 + 22 + 24 = - 55$
Mean $= 575 + 50 \times \frac{- 55}{110} = 575 - 25 = 550$
Modal class 550-600: l = 550, $f_{1} = 23$ , $f_{0} = 20$ , $f_{2} = 22$ , h = 50
Mode $= 550 + \frac{23 - 20}{2 ( 23 ) - 20 - 22} \times 50 = 550 + \frac{3}{4} \times 50 = 587.5$

Q35 (OR) (OR)5 marksLong AnswerStatistics CBSE 2026 · Standard 30/5/2

If the median of the following frequency distribution is 32.5 and sum of all frequencies is 40, then find the values of $f_{1}$ and $f_{2}$ :

Class Interval :	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequency :	3	$f_{1}$	9	12	6	$f_{2}$	2

Show answer & solution

Answer:

f_{1} = 5

f_{2} = 3

$3 + f_{1} + 9 + 12 + 6 + f_{2} + 2 = 40$ , so $f_{1} + f_{2} = 8$
Median 32.5 lies in 30-40: l = 30, f = 12, h = 10, cf $= 3 + f_{1} + 9 = 12 + f_{1}$ , $\frac{N}{2} = 20$
$32.5 = 30 + \frac{20 - ( 12 + f _{1} )}{12} \times 10$
$2.5 = \frac{( 8 - f _{1} ) \times 10}{12}$ , so $8 - f_{1} = 3$ and $f_{1} = 5$
$f_{2} = 8 - 5 = 3$

Q335 marksLong AnswerStatistics CBSE 2026 · Standard 30/5/3

The mean of the following distribution is 53. Find the missing frequency p.

Class Interval :	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100
Frequency :	12	15	p	28	13

Hence, find mode of the distribution.

Show answer & solution

Answer: p = 32; mode ≈ 56.19

Class marks: 10, 30, 50, 70, 90. $\sum f_{i} = 68 + p$
$\sum f_{i} x_{i} = 120 + 450 + 50 p + 1960 + 1170 = 3700 + 50 p$
$\frac{3700 + 50 p}{68 + p} = 53$
$3700 + 50 p = 3604 + 53 p$ , so $3 p = 96$ and $p = 32$
Modal class 40 – 60 (frequency 32): l = 40, $f_{1} = 32$ , $f_{0} = 15$ , $f_{2} = 28$ , h = 20
Mode $= 40 + \frac{32 - 15}{2 ( 32 ) - 15 - 28} \times 20 = 40 + \frac{17}{21} \times 20 = 40 + 16.19 = 56.19$ (approx.)

Q33 (OR) (OR)5 marksLong AnswerStatistics CBSE 2026 · Standard 30/5/3

Compute median of the following data :

Mid-value :	115	125	135	145	155	165	175
Frequency :	12	15	20	16	10	16	11

Show answer & solution

Answer: Median = 141.875

Class width = 10, so the classes are 110-120, 120-130, ..., 170-180.
N = 12 + 15 + 20 + 16 + 10 + 16 + 11 = 100, $\frac{N}{2} = 50$
Cumulative frequencies: 12, 27, 47, 63, 73, 89, 100
Median class 140-150: l = 140, cf = 47, f = 16, h = 10
Median $= 140 + \frac{50 - 47}{16} \times 10 = 140 + 1.875 = 141.875$

Q355 marksLong AnswerStatistics CBSE 2025 · Basic 430/1/1

The following data gives the information on the observed lifetime (in hours) of 200 electrical components :
Lifetime (in hours): 0 – 20, 20 – 40, 40 – 60, 60 – 80, 80 – 100, 100 – 120
Number of electrical components: 10, 35, 50, 60, 30, 15
Find the mean lifetime (in hours) of the electrical components.

Show answer & solution

Answer: 61 hours

Class marks $x_{i}$ : 10, 30, 50, 70, 90, 110.
$f_{i} x_{i}$ : 100, 1050, 2500, 4200, 2700, 1650.
$\sum f_{i} = 200$ , $\sum f_{i} x_{i} = 12200$ .
Mean $= \frac{12200}{200} = 61$ hours.

Q345 marksLong AnswerStatistics CBSE 2025 · Basic 430/1/2

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the monthly mean consumption from the data.
Monthly Consumption (in units): 50 – 100, 100 – 150, 150 – 200, 200 – 250, 250 – 300, 300 – 350, 350 – 400
Number of Consumers: 4, 5, 13, 20, 14, 8, 4

Show answer & solution

Answer: 230.15 units (approx.)

Class marks $x_{i}$ : 75, 125, 175, 225, 275, 325, 375. Take $a = 225$ , $h = 50$ , $u_{i} = \frac{x _{i} - 225}{50}$ .
$u_{i}$ : $- 3, - 2, - 1, 0, 1, 2, 3$ ; $f_{i} u_{i}$ : $- 12, - 10, - 13, 0, 14, 16, 12$ .
$\sum f_{i} = 68$ , $\sum f_{i} u_{i} = 7$ .
Mean $= 225 + 50 \times \frac{7}{68} = 225 + 5.15 = 230.15$ units.

Q325 marksLong AnswerStatistics CBSE 2025 · Basic 430/1/3

A life insurance agent found the following data for the distribution of 100 policy holders on the basis of their ages.
Age (in years): 15 – 20, 20 – 25, 25 – 30, 30 – 35, 35 – 40, 40 – 45, 45 – 50, 50 – 55, 55 – 60
Number of policy holders: 2, 4, 18, 21, 33, 11, 3, 6, 2
Find the median age of the policy holders.

Show answer & solution

Answer:

35 \frac{25}{33}

years ≈ 35.76 years

Cumulative frequencies: 2, 6, 24, 45, 78, 89, 92, 98, 100; $n = 100$ , $\frac{n}{2} = 50$ .
50 lies in the class 35 – 40, so it is the median class: $l = 35$ , $c f = 45$ , $f = 33$ , $h = 5$ .
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 35 + \frac{50 - 45}{33} \times 5 = 35 + \frac{25}{33} \approx 35.76$ years.

Q355 marksLong AnswerStatistics CBSE 2025 · Basic 430/2/1

The lengths of 40 leaves of a plant are measured, correct to the nearest millimetre and data obtained is represented in the following table :
Length in (mm): 100–120, 120–140, 140–160, 160–180, 180–200
Number of leaves: 8, 9, 12, 5, 6
Find the median length (in mm) of the leaves.

Show answer & solution

Answer: Median length = 145 mm

Cumulative frequencies: 8, 17, 29, 34, 40. n = 40, $\frac{n}{2} = 20$ .
cf just greater than 20 is 29, so the median class is 140–160.
$l = 140$ , cf = 17, f = 12, h = 20.
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 140 + \frac{20 - 17}{12} \times 20 = 140 + 5 = 145$ mm.

Also asked in: 2025 Basic 430/2/2, 2025 Basic 430/2/3

Q35 (OR) (OR)5 marksLong AnswerStatistics CBSE 2025 · Basic 430/2/1

A class teacher has the following absentees record of 30 students of a class.
Number of days: 0–4, 4–8, 8–12, 12–16, 16–20, 20–24
Number of Absent students: 1, 8, $x$ , 6, 5, $y$
If the mean number of days a student was absent is 12, find the values of $x$ and $y$ .

Show answer & solution

Answer:

x = 7

y = 3

Total: $1 + 8 + x + 6 + 5 + y = 30 \Rightarrow x + y = 10$ .
Class marks: 2, 6, 10, 14, 18, 22.
$\sum f x_{i} = 2 + 48 + 10 x + 84 + 90 + 22 y = 224 + 10 x + 22 y$ .
Mean $= 12 \Rightarrow 224 + 10 x + 22 y = 360 \Rightarrow 10 x + 22 y = 136 \Rightarrow 5 x + 11 y = 68$ .
Put $x = 10 - y$ : $50 - 5 y + 11 y = 68 \Rightarrow 6 y = 18 \Rightarrow y = 3$ , so $x = 7$ .

Also asked in: 2025 Basic 430/2/2, 2025 Basic 430/2/3

Q355 marksLong AnswerStatistics CBSE 2025 · Basic 430/3/1

The weights of 30 students of a class are given in the following distribution table :
Weight (in kg): 40 – 45, 45 – 50, 50 – 55, 55 – 60, 60 – 65, 65 – 70
Number of students: 2, 5, 8, 6, 6, 3
Find the median weight of the students.

Show answer & solution

Answer: 55 kg

Cumulative frequencies: 2, 7, 15, 21, 27, 30. $n = 30$ , $\frac{n}{2} = 15$ .
The cumulative frequency first reaches 15 in the class 50 – 55, so the median class is 50 – 55.
$l = 50$ , $c f = 7$ , $f = 8$ , $h = 5$ .
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 50 + \frac{15 - 7}{8} \times 5 = 50 + 5 = 55$ kg.

Q345 marksLong AnswerStatistics CBSE 2025 · Basic 430/3/2

The following distribution shows the weekly pocket allowance (in ₹) of some children of a locality. The mean pocket allowance is ₹ 180.
Weekly Pocket Allowance (in ₹): 110 – 130, 130 – 150, 150 – 170, 170 – 190, 190 – 210, 210 – 230, 230 – 250
Number of Children: 7, 6, 9, 13, f, 5, 4
Find the value of f. Hence find the mode of given data.

Show answer & solution

Answer:

f = 20

; mode

= ₹ 196 \frac{4}{11} \approx ₹ 196.36

Class marks: 120, 140, 160, 180, 200, 220, 240.
$\sum f_{i} = 44 + f$ and $\sum f_{i} x_{i} = 840 + 840 + 1440 + 2340 + 200 f + 1100 + 960 = 7520 + 200 f$ .
Mean: $\frac{7520 + 200 f}{44 + f} = 180 \Rightarrow 7520 + 200 f = 7920 + 180 f \Rightarrow 20 f = 400 \Rightarrow f = 20$ .
Highest frequency 20 is for 190 – 210, so the modal class is 190 – 210.
$l = 190$ , $f_{1} = 20$ , $f_{0} = 13$ , $f_{2} = 5$ , $h = 20$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 190 + \frac{7}{40 - 13 - 5} \times 20 = 190 + \frac{140}{22} \approx 196.36$ .
Mode $\approx ₹ 196.36$ .

Q325 marksLong AnswerStatistics CBSE 2025 · Basic 430/3/3

The following table shows the daily expenditure of 25 households of a locality.
Daily Expenditure (in ₹): 500 – 750, 750 – 1000, 1000 – 1250, 1250 – 1500, 1500 – 1750
Number of Households: 4, 2x + 1, 12, x, 2
Find the value of x. Hence find the mean daily expenditure.

Show answer & solution

Answer:

x = 2

; mean daily expenditure = ₹ 1055

$4 + (2 x + 1) + 12 + x + 2 = 25 \Rightarrow 3 x + 19 = 25 \Rightarrow x = 2$ .
Frequencies: 4, 5, 12, 2, 2. Class marks: 625, 875, 1125, 1375, 1625.
$\sum f_{i} x_{i} = 2500 + 4375 + 13500 + 2750 + 3250 = 26375$ .
Mean $= \frac{26375}{25} = 1055$ .
Mean daily expenditure = ₹ 1055.

Q335 marksLong AnswerStatistics CBSE 2025 · Basic 430/4/1

Find ‘mean’ and ‘mode’ of the following data :
Class: 15-20, 20-25, 25-30, 30-35, 35-40, 40-45
Frequency: 6, 16, 17, 4, 5, 2

Show answer & solution

Answer: Mean = 26.7; Mode =

25 \frac{5}{14} \approx 25.36

Class marks $x_{i}$ : 17.5, 22.5, 27.5, 32.5, 37.5, 42.5; $\sum f_{i} = 50$ .
$f_{i} x_{i}$ : 105, 360, 467.5, 130, 187.5, 85; $\sum f_{i} x_{i} = 1335$ .
Mean $= \frac{1335}{50} = 26.7$ .
Modal class 25-30: $l = 25$ , $f_{1} = 17$ , $f_{0} = 16$ , $f_{2} = 4$ , $h = 5$ .
Mode $= 25 + \frac{17 - 16}{2 \times 17 - 16 - 4} \times 5 = 25 + \frac{5}{14} \approx 25.36$ .

Q355 marksLong AnswerStatistics CBSE 2025 · Basic 430/4/2

Find ‘median’ and ‘mode’ of the following data :
Class: 100-105, 105-110, 110-115, 115-120, 120-125, 125-130
Frequency: 6, 8, 10, 4, 9, 3

Show answer & solution

Answer: Median = 113; Mode = 111.25

$N = 40$ ; cumulative frequencies: 6, 14, 24, 28, 37, 40.
$\frac{N}{2} = 20$ , so the median class is 110-115: $l = 110$ , $c f = 14$ , $f = 10$ , $h = 5$ .
Median $= 110 + \frac{20 - 14}{10} \times 5 = 113$ .
Modal class 110-115: $f_{1} = 10$ , $f_{0} = 8$ , $f_{2} = 4$ .
Mode $= 110 + \frac{10 - 8}{2 \times 10 - 8 - 4} \times 5 = 110 + \frac{10}{8} = 111.25$ .

Q355 marksLong AnswerStatistics CBSE 2025 · Basic 430/4/3

Find ‘mean’ and ‘mode’ of the following data :
Class: 20-25, 25-30, 30-35, 35-40, 40-45, 45-50
Frequency: 9, 8, 11, 13, 4, 5

Show answer & solution

Answer: Mean = 33.5; Mode =

35 \frac{10}{11} \approx 35.91

Class marks $x_{i}$ : 22.5, 27.5, 32.5, 37.5, 42.5, 47.5; $\sum f_{i} = 50$ .
$f_{i} x_{i}$ : 202.5, 220, 357.5, 487.5, 170, 237.5; $\sum f_{i} x_{i} = 1675$ .
Mean $= \frac{1675}{50} = 33.5$ .
Modal class 35-40: $l = 35$ , $f_{1} = 13$ , $f_{0} = 11$ , $f_{2} = 4$ , $h = 5$ .
Mode $= 35 + \frac{13 - 11}{2 \times 13 - 11 - 4} \times 5 = 35 + \frac{10}{11} \approx 35.91$ .

Q355 marksLong AnswerStatistics CBSE 2025 · Basic 430/5/1

Find 'mean' and 'mode' of the following data :
Class: 10-25, 25-40, 40-55, 55-70, 70-85, 85-100
Number of Students: 12, 10, 15, 13, 8, 12

Show answer & solution

Answer: Mean

\approx 54.14

; Mode

\approx 50.71

Class marks: 17.5, 32.5, 47.5, 62.5, 77.5, 92.5; take $a = 47.5$ , $h = 15$ , $u = \frac{x - 47.5}{15}$ : –2, –1, 0, 1, 2, 3.
$f u$ : –24, –10, 0, 13, 16, 36; $\sum f = 70$ , $\sum f u = 31$ .
Mean $= 47.5 + 15 \times \frac{31}{70} = 47.5 + 6.64 = 54.14$ (approx.).
Modal class 40-55: $l = 40$ , $f_{1} = 15$ , $f_{0} = 10$ , $f_{2} = 13$ , $h = 15$ .
Mode $= 40 + \frac{15 - 10}{30 - 10 - 13} \times 15 = 40 + \frac{75}{7} = 50.71$ (approx.).

Also asked in: 2025 Basic 430/5/2, 2025 Basic 430/5/3

Q35 (OR) (OR)5 marksLong AnswerStatistics CBSE 2025 · Basic 430/5/1

The following table shows the ages of patients admitted in a hospital during a year :
Age (in years): 5-15, 15-25, 25-35, 35-45, 45-55, 55-65
Number of Patients: 7, 10, 21, 22, 15, 5
Find 'mode' and 'median' of the above data.

Show answer & solution

Answer: Mode = 36.25 years; Median

\approx 35.91

years

Modal class 35-45: $l = 35$ , $f_{1} = 22$ , $f_{0} = 21$ , $f_{2} = 15$ , $h = 10$ .
Mode $= 35 + \frac{22 - 21}{44 - 21 - 15} \times 10 = 35 + \frac{10}{8} = 36.25$ .
Cumulative frequencies: 7, 17, 38, 60, 75, 80; $n = 80$ , $\frac{n}{2} = 40$ .
Median class 35-45: $l = 35$ , $c f = 38$ , $f = 22$ , $h = 10$ .
Median $= 35 + \frac{40 - 38}{22} \times 10 = 35 + 0.91 = 35.91$ (approx.).

Also asked in: 2025 Basic 430/5/2, 2025 Basic 430/5/3

Q355 marksLong AnswerStatistics CBSE 2025 · Basic 430/6/1

Find ‘mean’ and ‘mode’ of the following data :
Class: 15 – 20, 20 – 25, 25 – 30, 30 – 35, 35 – 40, 40 – 45
Frequency: 12, 10, 15, 11, 7, 5

Show answer & solution

Answer: Mean = 28; Mode

= \frac{250}{9} \approx 27.78

Class marks: 17.5, 22.5, 27.5, 32.5, 37.5, 42.5; $\sum f = 60$ .
$\sum f x = 210 + 225 + 412.5 + 357.5 + 262.5 + 212.5 = 1680$ .
Mean $= \frac{1680}{60} = 28$ .
Modal class 25 – 30 (highest frequency 15): $l = 25$ , $f_{1} = 15$ , $f_{0} = 10$ , $f_{2} = 11$ , $h = 5$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 25 + \frac{5}{30 - 10 - 11} \times 5 = 25 + \frac{25}{9} = \frac{250}{9} \approx 27.78$ .

Q345 marksLong AnswerStatistics CBSE 2025 · Basic 430/6/2

Find ‘mean’ and ‘mode’ of the following data :
Class: 0 – 15, 15 – 30, 30 – 45, 45 – 60, 60 – 75, 75 – 90
Frequency: 11, 8, 15, 7, 10, 9

Show answer & solution

Answer: Mean = 43.5; Mode = 37

Class marks: 7.5, 22.5, 37.5, 52.5, 67.5, 82.5; $\sum f = 60$ .
$\sum f x = 82.5 + 180 + 562.5 + 367.5 + 675 + 742.5 = 2610$ .
Mean $= \frac{2610}{60} = 43.5$ .
Modal class 30 – 45 (highest frequency 15): $l = 30$ , $f_{1} = 15$ , $f_{0} = 8$ , $f_{2} = 7$ , $h = 15$ .
Mode $= 30 + \frac{15 - 8}{30 - 8 - 7} \times 15 = 30 + \frac{7}{15} \times 15 = 37$ .

Q325 marksLong AnswerStatistics CBSE 2025 · Basic 430/6/3

Find ‘mean’ and ‘mode’ marks of the following data :
Marks: 0 – 5, 5 – 10, 10 – 15, 15 – 20, 20 – 25, 25 – 30
Number of students: 2, 3, 8, 15, 14, 8

Show answer & solution

Answer: Mean = 18.5; Mode = 19.375

Class marks: 2.5, 7.5, 12.5, 17.5, 22.5, 27.5; $\sum f = 50$ .
$\sum f x = 5 + 22.5 + 100 + 262.5 + 315 + 220 = 925$ .
Mean $= \frac{925}{50} = 18.5$ .
Modal class 15 – 20 (highest frequency 15): $l = 15$ , $f_{1} = 15$ , $f_{0} = 8$ , $f_{2} = 14$ , $h = 5$ .
Mode $= 15 + \frac{15 - 8}{30 - 8 - 14} \times 5 = 15 + \frac{35}{8} = 19.375$ .

Q355 marksLong AnswerStatistics CBSE 2025 · Standard 30/1/1

Find the missing frequency 'f' in the following table, if the mean of the given data is 18. Hence find the mode.
Daily Allowance: 11 – 13, 13 – 15, 15 – 17, 17 – 19, 19 – 21, 21 – 23, 23 – 25
Number of Children: 7, 6, 9, 13, f, 5, 4

Show answer & solution

Answer:

f = 20

; Mode

= 19 \frac{7}{11} \approx 19.64

Class marks $x_{i}$ : 12, 14, 16, 18, 20, 22, 24.
$f_{i} x_{i}$ : 84, 84, 144, 234, $20 f$ , 110, 96.
$\sum f_{i} = 44 + f$ and $\sum f_{i} x_{i} = 752 + 20 f$ .
Mean $= \frac{752 + 20 f}{44 + f} = 18$
$752 + 20 f = 792 + 18 f$ , so $2 f = 40$ and $f = 20$ .
Mode: the highest frequency is 20, so the modal class is 19 – 21.
$l = 19$ , $f_{1} = 20$ , $f_{0} = 13$ , $f_{2} = 5$ , $h = 2$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 19 + \frac{7}{40 - 13 - 5} \times 2 = 19 + \frac{14}{22} = 19 + \frac{7}{11}$
Mode $= 19 \frac{7}{11} \approx 19.64$ .

Q335 marksLong AnswerStatistics CBSE 2025 · Standard 30/1/2

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the mean and mode of the data :
Monthly Consumption (in units): 65 – 85, 85 – 105, 105 – 125, 125 – 145, 145 – 165, 165 – 185, 185 – 205
Number of Consumers: 4, 5, 13, 20, 14, 8, 4

Show answer & solution

Answer: Mean

= 137 \frac{1}{17} \approx 137.06

units; Mode

= 135 \frac{10}{13} \approx 135.77

units

Class marks $x_{i}$ : 75, 95, 115, 135, 155, 175, 195. Take assumed mean $a = 135$ , $h = 20$ , $u_{i} = \frac{x _{i} - 135}{20}$ : $- 3, - 2, - 1, 0, 1, 2, 3$ .
$f_{i} u_{i}$ : $- 12, - 10, - 13, 0, 14, 16, 12$ ; $\sum f_{i} u_{i} = 7$ , $\sum f_{i} = 68$ .
Mean $= 135 + \frac{7}{68} \times 20 = 135 + \frac{35}{17} \approx 135 + 2.06 = 137.06$ units.
Mode: the highest frequency is 20, so the modal class is 125 – 145.
$l = 125$ , $f_{1} = 20$ , $f_{0} = 13$ , $f_{2} = 14$ , $h = 20$ .
Mode $= 125 + \frac{20 - 13}{40 - 13 - 14} \times 20 = 125 + \frac{140}{13} \approx 125 + 10.77 = 135.77$ units.

Q345 marksLong AnswerStatistics CBSE 2025 · Standard 30/1/3

The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :
Length (in mm): 118 – 126, 127 – 135, 136 – 144, 145 – 153, 154 – 162, 163 – 171, 172 – 180
Number of Leaves: 3, 5, 9, 12, 5, 4, 2
Find the median length of the leaves.

Show answer & solution

Answer:

146.75

The classes are not continuous; subtract 0.5 from lower limits and add 0.5 to upper limits:
117.5 – 126.5, 126.5 – 135.5, 135.5 – 144.5, 144.5 – 153.5, 153.5 – 162.5, 162.5 – 171.5, 171.5 – 180.5.
Cumulative frequencies: 3, 8, 17, 29, 34, 38, 40. $n = 40$ , $\frac{n}{2} = 20$ .
The median class is 144.5 – 153.5 (cumulative frequency 29 first exceeds 20).
$l = 144.5$ , $c f = 17$ , $f = 12$ , $h = 9$ .
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 144.5 + \frac{20 - 17}{12} \times 9 = 144.5 + 2.25 = 146.75$ mm.

Q355 marksLong AnswerStatistics CBSE 2025 · Standard 30/2/1

Find the Mean and Mode of the following frequency distribution :
Class: 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60, 60 – 70
Frequency: 8, 7, 15, 20, 12, 8, 10

Show answer & solution

Answer: Mean

= 35.625

; Mode

= \frac{440}{13} \approx 33.85

Class marks $x_{i}$ : 5, 15, 25, 35, 45, 55, 65
$f_{i} x_{i}$ : 40, 105, 375, 700, 540, 440, 650
$\sum f_{i} = 80$ , $\sum f_{i} x_{i} = 2850$
Mean $= \frac{2850}{80} = 35.625$
Modal class is 30 – 40 (highest frequency 20): $l = 30$ , $f_{1} = 20$ , $f_{0} = 15$ , $f_{2} = 12$ , $h = 10$
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 30 + \frac{5}{13} \times 10 = 30 + \frac{50}{13} \approx 33.85$

Q345 marksLong AnswerStatistics CBSE 2025 · Standard 30/2/2

Find the mean and median for the following data :
Classes: 5 – 15, 15 – 25, 25 – 35, 35 – 45, 45 – 55, 55 – 65, 65 – 75
Frequency: 2, 3, 5, 7, 4, 2, 2

Show answer & solution

Answer: Mean

= 38.8

; Median

= \frac{270}{7} \approx 38.57

Class marks $x_{i}$ : 10, 20, 30, 40, 50, 60, 70
$f_{i} x_{i}$ : 20, 60, 150, 280, 200, 120, 140
$\sum f_{i} = 25$ , $\sum f_{i} x_{i} = 970$
Mean $= \frac{970}{25} = 38.8$
Cumulative frequencies: 2, 5, 10, 17, 21, 23, 25; $\frac{N}{2} = 12.5$
Median class is 35 – 45: $l = 35$ , $c f = 10$ , $f = 7$ , $h = 10$
Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 35 + \frac{2.5}{7} \times 10 = 35 + \frac{25}{7} = \frac{270}{7} \approx 38.57$

Q345 marksLong AnswerStatistics CBSE 2025 · Standard 30/2/3

Find the Mean and Mode of the following data :
Class: 4 – 8, 8 – 12, 12 – 16, 16 – 20, 20 – 24, 24 – 28, 28 – 32, 32 – 36
Frequency: 2, 12, 15, 25, 18, 12, 13, 3

Show answer & solution

Answer: Mean

= 19.92

; Mode

= \frac{312}{17} \approx 18.35

Class marks $x_{i}$ : 6, 10, 14, 18, 22, 26, 30, 34
$f_{i} x_{i}$ : 12, 120, 210, 450, 396, 312, 390, 102
$\sum f_{i} = 100$ , $\sum f_{i} x_{i} = 1992$
Mean $= \frac{1992}{100} = 19.92$
Modal class is 16 – 20 (highest frequency 25): $l = 16$ , $f_{1} = 25$ , $f_{0} = 15$ , $f_{2} = 18$ , $h = 4$
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 16 + \frac{10}{17} \times 4 = 16 + \frac{40}{17} = \frac{312}{17} \approx 18.35$

Q355 marksLong AnswerStatistics CBSE 2025 · Standard 30/4/1

The following table shows the number of patients of different age group who were discharged from the hospital in a particular month :
Age (in years): 5-15, 15-25, 25-35, 35-45, 45-55, 55-65, Total
Number of Patients Discharged: 6, 11, 21, 23, 14, 5, 80
Find the 'mean' and the 'mode' of the above data.

Show answer & solution

Answer: Mean

= 35.375

years; Mode

= 35 + \frac{20}{11} \approx 36.82

years

Class marks: 10, 20, 30, 40, 50, 60
$\sum f_{i} x_{i} = 60 + 220 + 630 + 920 + 700 + 300 = 2830$ ; $\sum f_{i} = 80$
Mean $= \frac{2830}{80} = 35.375$
Modal class 35-45: $l = 35$ , $f_{1} = 23$ , $f_{0} = 21$ , $f_{2} = 14$ , $h = 10$
Mode $= 35 + \frac{23 - 21}{2 ( 23 ) - 21 - 14} \times 10 = 35 + \frac{20}{11} \approx 36.82$

Q325 marksLong AnswerStatistics CBSE 2025 · Standard 30/4/2

The following table shows the number of traffic challans issued in the month of April by the traffic police :
Number of Challans: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, Total
Number of Days: 3, 5, 10, 9, 2, 1, 30
Find the 'mean' and 'mode' of the above data.

Show answer & solution

Answer: Mean

= \frac{80}{3} \approx 26.67

; Mode

= \frac{85}{3} \approx 28.33

Class marks: 5, 15, 25, 35, 45, 55
$\sum f_{i} x_{i} = 15 + 75 + 250 + 315 + 90 + 55 = 800$ ; $\sum f_{i} = 30$
Mean $= \frac{800}{30} = \frac{80}{3} \approx 26.67$
Modal class 20-30: $l = 20$ , $f_{1} = 10$ , $f_{0} = 5$ , $f_{2} = 9$ , $h = 10$
Mode $= 20 + \frac{10 - 5}{20 - 5 - 9} \times 10 = 20 + \frac{50}{6} = \frac{85}{3} \approx 28.33$

Q345 marksLong AnswerStatistics CBSE 2025 · Standard 30/4/3

Following table shows the absentees record of 40 students in an academic year :
Number of Days: 2-6, 6-10, 10-14, 14-18, 18-22, 22-26, 26-30
Number of Students: 11, 10, 7, 4, 4, 3, 1
Find the 'mean' and the 'mode' of the above data.

Show answer & solution

Answer: Mean

= 11.3

days; Mode

= \frac{17}{3} \approx 5.67

days

Class marks: 4, 8, 12, 16, 20, 24, 28
$\sum f_{i} x_{i} = 44 + 80 + 84 + 64 + 80 + 72 + 28 = 452$ ; $\sum f_{i} = 40$
Mean $= \frac{452}{40} = 11.3$
Modal class 2-6: $l = 2$ , $f_{1} = 11$ , $f_{0} = 0$ , $f_{2} = 10$ , $h = 4$
Mode $= 2 + \frac{11 - 0}{22 - 0 - 10} \times 4 = 2 + \frac{11}{3} = \frac{17}{3} \approx 5.67$

Q355 marksLong AnswerStatistics CBSE 2025 · Standard 30/5/1

Medical check-up was carried out for 35 students of a class and their weights were recorded as follows :
Weight (in kg): 38-40, 40-42, 42-44, 44-46, 46-48, 48-50, 50-52
Number of Students: 3, 2, 4, 5, 14, 4, 3
Find the difference between the mean weight and the median weight.

Show answer & solution

Answer: Mean = 45.8 kg, median = 46.5 kg; difference = 0.7 kg

Class marks: 39, 41, 43, 45, 47, 49, 51
$\sum f x = 117 + 82 + 172 + 225 + 658 + 196 + 153 = 1603$ , $\sum f = 35$
Mean $= \frac{1603}{35} = 45.8$ kg
Cumulative frequencies: 3, 5, 9, 14, 28, 32, 35; $\frac{n}{2} = 17.5$ , so median class is 46-48.
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 46 + \frac{17.5 - 14}{14} \times 2 = 46 + 0.5 = 46.5$ kg
Difference $= 46.5 - 45.8 = 0.7$ kg

Q325 marksLong AnswerStatistics CBSE 2025 · Standard 30/5/2

During a medical checkup, height of 35 students of a class were recorded as follows :
Height (in cm): 90-100, 100-110, 110-120, 120-130, 130-140, 140-150
Number of Students: 3, 2, 4, 5, 14, 7
Find the difference between the mean height and median height.

Show answer & solution

Answer: Mean

= \frac{897}{7} \approx 128.14

cm, median

= 132.5

cm; difference

= \frac{61}{14} \approx 4.36

Class marks: 95, 105, 115, 125, 135, 145
$\sum f x = 285 + 210 + 460 + 625 + 1890 + 1015 = 4485$ , $\sum f = 35$
Mean $= \frac{4485}{35} = \frac{897}{7} \approx 128.14$ cm
Cumulative frequencies: 3, 5, 9, 14, 28, 35; $\frac{n}{2} = 17.5$ , so median class is 130-140.
Median $= 130 + \frac{17.5 - 14}{14} \times 10 = 130 + 2.5 = 132.5$ cm
Difference $= 132.5 - \frac{897}{7} = \frac{61}{14} \approx 4.36$ cm

Q345 marksLong AnswerStatistics CBSE 2025 · Standard 30/5/3

The following table gives the daily income of 50 cab drivers of a particular city :
Income (₹): 500 - 600, 600 - 700, 700 - 800, 800 - 900, 900 - 1000
No. of Drivers: 12, 14, 8, 6, 10
Find the mean income and the modal income.

Show answer & solution

Answer: Mean income = ₹ 726; modal income = ₹ 625

Class marks: 550, 650, 750, 850, 950
$\sum f x = 6600 + 9100 + 6000 + 5100 + 9500 = 36300$ , $\sum f = 50$
Mean $= \frac{36300}{50} = 726$ , i.e. ₹ 726
Modal class: 600 - 700 ( $f_{1} = 14$ , $f_{0} = 12$ , $f_{2} = 8$ , $l = 600$ , $h = 100$ )
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 600 + \frac{2}{8} \times 100 = 625$ , i.e. ₹ 625

Q355 marksLong AnswerStatistics CBSE 2025 · Standard 30/6/1

Following distribution shows the marks of 230 students in a particular subject. If the median marks are 46, then find the values of $x$ and y.
Marks: 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60, 60 – 70, 70 – 80
Number of Students: 12, 30, $x$ , 65, y, 25, 18

Show answer & solution

Answer:

x = 34

y = 46

Total: $12 + 30 + x + 65 + y + 25 + 18 = 230 \Rightarrow x + y = 80$
$\frac{N}{2} = 115$ ; median 46 lies in class 40 – 50: $l = 40$ , $f = 65$ , $h = 10$ , $c f = 42 + x$
$46 = 40 + \frac{115 - ( 42 + x )}{65} \times 10$
$6 = \frac{( 73 - x ) \times 10}{65} \Rightarrow 73 - x = 39 \Rightarrow x = 34$
$y = 80 - 34 = 46$

Q325 marksLong AnswerStatistics CBSE 2025 · Standard 30/6/2

Following data shows the number of family members living in different bungalows of a locality :
Number of Members: 0 – 2, 2 – 4, 4 – 6, 6 – 8, 8 – 10, Total
Number of Bungalows: 10, p, 60, q, 5, 120
If the median number of members is found to be 5, find the values of p and q.

Show answer & solution

Answer:

p = 20

q = 25

$10 + p + 60 + q + 5 = 120 \Rightarrow p + q = 45$
$\frac{N}{2} = 60$ ; median 5 lies in class 4 – 6: $l = 4$ , $f = 60$ , $h = 2$ , $c f = 10 + p$
$5 = 4 + \frac{60 - ( 10 + p )}{60} \times 2$
$1 = \frac{50 - p}{30} \Rightarrow p = 20$
$q = 45 - 20 = 25$

Q345 marksLong AnswerStatistics CBSE 2025 · Standard 30/6/3

The population of lions was noted in different regions across the world in the following table :
Number of lions: 0 – 100, 100 – 200, 200 – 300, 300 – 400, 400 – 500, 500 – 600, 600 – 700, 700 – 800, 800 – 900, 900 – 1000
Number of regions: 2, 5, 9, 12, $x$ , 20, 15, 9, y, 2
Total: 100
If the median of the given data is 525, find the values of $x$ and y.

Show answer & solution

Answer:

x = 17

y = 9

$2 + 5 + 9 + 12 + x + 20 + 15 + 9 + y + 2 = 100 \Rightarrow x + y = 26$
$\frac{N}{2} = 50$ ; median 525 lies in class 500 – 600: $l = 500$ , $f = 20$ , $h = 100$ , $c f = 28 + x$
$525 = 500 + \frac{50 - ( 28 + x )}{20} \times 100$
$25 = 5 (22 - x) \Rightarrow 22 - x = 5 \Rightarrow x = 17$
$y = 26 - 17 = 9$

Q325 marksLong AnswerStatistics CBSE 2024 · Basic 430/3/1

Consider the following distribution of hourly wages of 50 workers of a factory :
Hourly wages (in ₹): 100-120, 120-140, 140-160, 160-180, 180-200
Number of workers: 12, 14, 8, 6, 10
Find the mean and the median of the above data.

Show answer & solution

Answer: Mean = ₹145.20; Median = ₹

\frac{970}{7} \approx

₹138.57

Class marks $x_{i}$ : 110, 130, 150, 170, 190; $f_{i}$ : 12, 14, 8, 6, 10; $\sum f_{i} = 50$ .
$f_{i} x_{i}$ : 1320, 1820, 1200, 1020, 1900; $\sum f_{i} x_{i} = 7260$ .
Mean $= \frac{7260}{50} = 145.2$ .
Cumulative frequencies: 12, 26, 34, 40, 50; $\frac{n}{2} = 25$ , so the median class is 120-140.
$l = 120$ , $c f = 12$ , $f = 14$ , $h = 20$ .
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 120 + \frac{25 - 12}{14} \times 20 = 120 + \frac{130}{7} \approx 138.57$ .

Q355 marksLong AnswerStatistics CBSE 2024 · Basic 430/3/2

Find the mean and median of the following distribution :
Class: 0 – 20, 20 – 40, 40 – 60, 60 – 80, 80 – 100, 100 – 120
Frequency: 5, 8, 10, 12, 7, 8

Show answer & solution

Answer: Mean = 62.8; Median

= \frac{190}{3} \approx 63.33

Class marks $x_{i}$ : 10, 30, 50, 70, 90, 110; $f_{i}$ : 5, 8, 10, 12, 7, 8; $\sum f_{i} = 50$ .
$f_{i} x_{i}$ : 50, 240, 500, 840, 630, 880; $\sum f_{i} x_{i} = 3140$ .
Mean $= \frac{3140}{50} = 62.8$ .
Cumulative frequencies: 5, 13, 23, 35, 42, 50; $\frac{n}{2} = 25$ , so the median class is 60 – 80.
$l = 60$ , $c f = 23$ , $f = 12$ , $h = 20$ .
Median $= 60 + \frac{25 - 23}{12} \times 20 = 60 + \frac{10}{3} \approx 63.33$ .

Q335 marksLong AnswerStatistics CBSE 2024 · Basic 430/3/3

The marks obtained by 45 students of a class in a test are given below :
Marks: 40 – 45, 45 – 50, 50 – 55, 55 – 60, 60 – 65, 65 – 70
No. of Students: 8, 9, 10, 9, 5, 4
Find the mean and median marks.

Show answer & solution

Answer: Mean

= \frac{2392.5}{45} \approx 53.17

; Median = 52.75

Class marks $x_{i}$ : 42.5, 47.5, 52.5, 57.5, 62.5, 67.5; $f_{i}$ : 8, 9, 10, 9, 5, 4; $\sum f_{i} = 45$ .
$f_{i} x_{i}$ : 340, 427.5, 525, 517.5, 312.5, 270; $\sum f_{i} x_{i} = 2392.5$ .
Mean $= \frac{2392.5}{45} \approx 53.17$ .
Cumulative frequencies: 8, 17, 27, 36, 41, 45; $\frac{n}{2} = 22.5$ , so the median class is 50 – 55.
$l = 50$ , $c f = 17$ , $f = 10$ , $h = 5$ .
Median $= 50 + \frac{22.5 - 17}{10} \times 5 = 50 + 2.75 = 52.75$ .

Q355 marksLong AnswerStatistics CBSE 2024 · Basic 430/5/1

The following frequency distribution table gives the monthly consumption of electricity of 70 consumers of a locality. Find the median of the data.
Monthly consumption (in units): 65 – 85, 85 – 105, 105 – 125, 125 – 145, 145 – 165, 165 – 185, 185 – 205
Number of consumers: 7, 8, 7, 20, 14, 9, 5

Show answer & solution

Answer: 138 units

Cumulative frequencies: 7, 15, 22, 42, 56, 65, 70; $N = 70$ .
$\frac{N}{2} = 35$ , which lies in the class 125 – 145 (median class).
$l = 125$ , $c f = 22$ , $f = 20$ , $h = 20$ .
Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 125 + \frac{35 - 22}{20} \times 20$ .
Median $= 125 + 13 = 138$ units.

Q335 marksLong AnswerStatistics CBSE 2024 · Basic 430/5/2

The following table shows the ages of the patients admitted in a hospital during a year :
Age (in years): 5 – 15, 15 – 25, 25 – 35, 35 – 45, 45 – 55, 55 – 65
Number of patients: 6, 11, 21, 23, 14, 5
Find the median of the above given data.

Show answer & solution

Answer:

35 + \frac{20}{23} \approx 35.87

years

Cumulative frequencies: 6, 17, 38, 61, 75, 80; $N = 80$ .
$\frac{N}{2} = 40$ , which lies in the class 35 – 45 (median class).
$l = 35$ , $c f = 38$ , $f = 23$ , $h = 10$ .
Median $= 35 + \frac{40 - 38}{23} \times 10 = 35 + \frac{20}{23}$ .
Median $\approx 35.87$ years.

Q335 marksLong AnswerStatistics CBSE 2024 · Basic 430/5/3

The following data gives the information about the lifetimes (in hours) of 225 neon lamps :
Lifetime (in hours): 1500 – 2000, 2000 – 2500, 2500 – 3000, 3000 – 3500, 3500 – 4000, 4000 – 4500
Number of lamps: 10, 35, 52, 61, 38, 29
Find the median lifetime of a lamp.

Show answer & solution

Answer:

3000 + \frac{7750}{61} \approx 3127.05

hours

Cumulative frequencies: 10, 45, 97, 158, 196, 225; $N = 225$ .
$\frac{N}{2} = 112.5$ , which lies in the class 3000 – 3500 (median class).
$l = 3000$ , $c f = 97$ , $f = 61$ , $h = 500$ .
Median $= 3000 + \frac{112.5 - 97}{61} \times 500 = 3000 + \frac{7750}{61}$ .
Median $\approx 3000 + 127.05 = 3127.05$ hours.

Q355 marksLong AnswerStatistics CBSE 2024 · Standard 30/4/1

The following table shows the ages of the patients admitted in a hospital during a year :
Age (in years): 5–15, 15–25, 25–35, 35–45, 45–55, 55–65
Number of patients: 6, 11, 21, 23, 14, 5
Find the mode and mean of the data given above.

Show answer & solution

Answer: Mode =

36 \frac{9}{11} \approx 36.82

years; Mean = 35.375

\approx

35.38 years

Mode: the modal class is 35–45 (highest frequency 23). $l = 35$ , $f_{1} = 23$ , $f_{0} = 21$ , $f_{2} = 14$ , $h = 10$ .
Mode = $l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 35 + \frac{2}{46 - 21 - 14} \times 10 = 35 + \frac{20}{11} \approx 36.82$ years.
Mean: class marks 10, 20, 30, 40, 50, 60; $f_{i} x_{i}$ = 60, 220, 630, 920, 700, 300.
$Σ f_{i} = 80$ , $Σ f_{i} x_{i} = 2830$ .
Mean = $\frac{2830}{80} = 35.375 \approx 35.38$ years.

Also asked in: 2024 Standard 30/4/2

Q325 marksLong AnswerStatistics CBSE 2024 · Standard 30/4/3

The following distribution shows the daily pocket allowance of children of a locality. The mean daily pocket allowance is ₹ 36.10. Find the missing frequency, f.
Daily pocket allowance (in ₹): 20–25, 25–30, 30–35, 35–40, 40–45, 45–50, 50–55
Number of children: 7, 6, 9, 13, f, 5, 4

Show answer & solution

Answer: f = 6

Class marks: 22.5, 27.5, 32.5, 37.5, 42.5, 47.5, 52.5.
$f_{i} x_{i}$ : 157.5, 165, 292.5, 487.5, 42.5f, 237.5, 210.
$Σ f_{i} = 44 + f$ and $Σ f_{i} x_{i} = 1550 + 42.5 f$ .
$\frac{1550 + 42.5 f}{44 + f} = 36.10$ , so $1550 + 42.5 f = 1588.4 + 36.1 f$ .
$6.4 f = 38.4$ , so $f = 6$ .

Q345 marksLong AnswerStatistics CBSE 2023 · Basic 430/1/1

The table given below shows the daily expenditure on food of 25 households in a locality :
Daily expenditure (₹): 100 – 150, 150 – 200, 200 – 250, 250 – 300, 300 – 350
Number of household: 4, 5, 12, 2, 2
Find the mean daily expenditure on food. Also, find the mode of the data.

Show answer & solution

Answer: Mean = ₹211; Mode = ₹

\frac{3750}{17}

≈ ₹220.59

Class marks $x_{i}$ : 125, 175, 225, 275, 325; take $a = 225$ , $h = 50$ , $u_{i} = \frac{x _{i} - 225}{50}$ : $- 2, - 1, 0, 1, 2$ .
$f_{i} u_{i}$ : $- 8, - 5, 0, 2, 4$ ; $\sum f_{i} = 25$ , $\sum f_{i} u_{i} = - 7$ .
Mean $= a + h \frac{\sum f _{i} u _{i}}{\sum f _{i}} = 225 + 50 \times \frac{- 7}{25} = 225 - 14 = 211$
Modal class is 200 – 250 (highest frequency 12): $l = 200$ , $f_{1} = 12$ , $f_{0} = 5$ , $f_{2} = 2$ , $h = 50$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 200 + \frac{7}{17} \times 50$
$= 200 + \frac{350}{17} \approx 220.59$

Also asked in: 2023 Basic 430/1/2

Q325 marksLong AnswerStatistics CBSE 2023 · Basic 430/1/3

A survey conducted on 20 families in a locality by a group of students resulted in the following frequency table for the number of family members in a family.
Family size: 1 – 3, 3 – 5, 5 – 7, 7 – 9, 9 – 11
Number of families: 7, 8, 2, 2, 1
Determine the mean and mode of the above data.

Show answer & solution

Answer: Mean = 4.2; Mode =

\frac{23}{7}

≈ 3.29

Class marks $x_{i}$ : 2, 4, 6, 8, 10; $f_{i}$ : 7, 8, 2, 2, 1; $\sum f_{i} = 20$ .
$f_{i} x_{i}$ : 14, 32, 12, 16, 10; $\sum f_{i} x_{i} = 84$ .
Mean $= \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{84}{20} = 4.2$
Modal class is 3 – 5 (highest frequency 8): $l = 3$ , $f_{1} = 8$ , $f_{0} = 7$ , $f_{2} = 2$ , $h = 2$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 3 + \frac{1}{16 - 7 - 2} \times 2$
$= 3 + \frac{2}{7} = \frac{23}{7} \approx 3.29$

Q335 marksLong AnswerStatistics CBSE 2023 · Basic 430/2/1

Find the mean and the median of the following data :
Marks: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60, 60–70, 70–80
Number of Students: 3, 5, 16, 12, 13, 20, 6, 5

Show answer & solution

Answer: Mean = 42; Median =

43 \frac{1}{13}

≈ 43.08

Class marks $x_{i}$ : 5, 15, 25, 35, 45, 55, 65, 75; $Σ f_{i} = 80$
$f_{i} x_{i}$ : 15, 75, 400, 420, 585, 1100, 390, 375; $Σ f_{i} x_{i} = 3360$
Mean $= \frac{3360}{80} = 42$
Cumulative frequencies: 3, 8, 24, 36, 49, 69, 75, 80
$\frac{N}{2} = 40$ , so the median class is 40–50 with l = 40, cf = 36, f = 13, h = 10
Median $= 40 + \frac{40 - 36}{13} \times 10 = 40 + \frac{40}{13} \approx 43.08$

Q345 marksLong AnswerStatistics CBSE 2023 · Basic 430/2/2

Find the mean and the median of the following data :
Class: 85–90, 90–95, 95–100, 100–105, 105–110, 110–115
Frequency: 10, 12, 15, 14, 12, 7

Show answer & solution

Answer: Mean ≈ 99.43; Median =

99 \frac{1}{3}

≈ 99.33

Class marks $x_{i}$ : 87.5, 92.5, 97.5, 102.5, 107.5, 112.5; $Σ f_{i} = 70$
$f_{i} x_{i}$ : 875, 1110, 1462.5, 1435, 1290, 787.5; $Σ f_{i} x_{i} = 6960$
Mean $= \frac{6960}{70} \approx 99.43$
Cumulative frequencies: 10, 22, 37, 51, 63, 70
$\frac{N}{2} = 35$ , so the median class is 95–100 with l = 95, cf = 22, f = 15, h = 5
Median $= 95 + \frac{35 - 22}{15} \times 5 = 95 + \frac{13}{3} \approx 99.33$

Q325 marksLong AnswerStatistics CBSE 2023 · Basic 430/2/3

Find the mean and the median of the marks of 100 students of a class, given in the following table :
Marks: 0–5, 5–10, 10–15, 15–20, 20–25, 25–30
Number of students: 4, 11, 13, 15, 31, 26

Show answer & solution

Answer: Mean = 19.3; Median =

21 \frac{4}{31}

≈ 21.13

Class marks $x_{i}$ : 2.5, 7.5, 12.5, 17.5, 22.5, 27.5; $Σ f_{i} = 100$
$f_{i} x_{i}$ : 10, 82.5, 162.5, 262.5, 697.5, 715; $Σ f_{i} x_{i} = 1930$
Mean $= \frac{1930}{100} = 19.3$
Cumulative frequencies: 4, 15, 28, 43, 74, 100
$\frac{N}{2} = 50$ , so the median class is 20–25 with l = 20, cf = 43, f = 31, h = 5
Median $= 20 + \frac{50 - 43}{31} \times 5 = 20 + \frac{35}{31} = 21 \frac{4}{31} \approx 21.13$

Q345 marksLong AnswerStatistics CBSE 2023 · Basic 430/4/1

The following table gives the monthly consumption of electricity of 100 families :
Monthly Consumption (in units): 130-140, 140-150, 150-160, 160-170, 170-180, 180-190, 190-200
Number of families: 5, 9, 17, 28, 24, 10, 7
Find the median of the above data.

Show answer & solution

Answer: Median

= 160 + \frac{95}{14} \approx 166.79

units

Cumulative frequencies: 5, 14, 31, 59, 83, 93, 100. $N = 100$ , $\frac{N}{2} = 50$ .
Median class is 160-170 (cf just above 50 is 59).
$l = 160$ , $c f = 31$ , $f = 28$ , $h = 10$ .
Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 160 + \frac{50 - 31}{28} \times 10$ .
$= 160 + \frac{190}{28} = 160 + 6.79 = 166.79$ units (approx).

Also asked in: 2023 Basic 430/4/3

Q335 marksLong AnswerStatistics CBSE 2023 · Basic 430/4/2

The distribution below gives the weights of 30 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, 70-75
Number of Students: 2, 3, 8, 6, 6, 3, 2

Show answer & solution

Answer: Median

= \frac{170}{3} \approx 56.67

Cumulative frequencies: 2, 5, 13, 19, 25, 28, 30. $N = 30$ , $\frac{N}{2} = 15$ .
Median class is 55-60.
$l = 55$ , $c f = 13$ , $f = 6$ , $h = 5$ .
Median $= 55 + \frac{15 - 13}{6} \times 5 = 55 + \frac{5}{3} = 56.67$ kg (approx).

Q345 marksLong AnswerStatistics CBSE 2023 · Standard 30/2/1

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mean and median of the following data.
Number of cars: 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60, 60 – 70, 70 – 80
Frequency (periods): 7, 14, 13, 12, 20, 11, 15, 8

Show answer & solution

Answer: Mean = 40.7, Median = 42

Class marks: 5, 15, 25, 35, 45, 55, 65, 75
$f_{i} x_{i}$ : 35, 210, 325, 420, 900, 605, 975, 600; $\sum f_{i} = 100$ , $\sum f_{i} x_{i} = 4070$
Mean $= \frac{4070}{100} = 40.7$
Cumulative frequencies: 7, 21, 34, 46, 66, 77, 92, 100; $\frac{N}{2} = 50$ , so median class is 40 – 50.
$l = 40$ , $c f = 46$ , $f = 20$ , $h = 10$
Median $= 40 + \frac{50 - 46}{20} \times 10 = 42$

Also asked in: 2023 Standard 30/2/2

Q325 marksLong AnswerStatistics CBSE 2023 · Standard 30/2/3

The mode of the following frequency distribution is 55. Find the missing frequencies ‘a’ and ‘b’.
Class Interval: 0 – 15, 15 – 30, 30 – 45, 45 – 60, 60 – 75, 75 – 90, Total
Frequency: 6, 7, a, 15, 10, b, 51

Show answer & solution

Answer:

a = 5

b = 8

Total: $6 + 7 + a + 15 + 10 + b = 51 \Rightarrow a + b = 13$
Mode 55 lies in 45 – 60, so $l = 45$ , $h = 15$ , $f_{1} = 15$ , $f_{0} = a$ , $f_{2} = 10$ .
$55 = 45 + \frac{15 - a}{2 ( 15 ) - a - 10} \times 15 \Rightarrow \frac{15 ( 15 - a )}{20 - a} = 10$
$225 - 15 a = 200 - 10 a \Rightarrow a = 5$
$b = 13 - 5 = 8$

Q355 marksLong AnswerStatistics CBSE 2023 · Standard 30/4/1

The monthly expenditure on milk in 200 families of a Housing Society is given below :
Monthly Expenditure (in ₹): 1000-1500, 1500-2000, 2000-2500, 2500-3000, 3000-3500, 3500-4000, 4000-4500, 4500-5000
Number of families: 24, 40, 33, $x$ , 30, 22, 16, 7
Find the value of $x$ and also, find the median and mean expenditure on milk.

Show answer & solution

Answer:

x = 28

; median ≈ ₹2553.57; mean = ₹2662.50

Total: $24 + 40 + 33 + x + 30 + 22 + 16 + 7 = 172 + x = 200 \Rightarrow x = 28$
Cumulative frequencies: 24, 64, 97, 125, 155, 177, 193, 200
$\frac{N}{2} = 100$ , so the median class is 2500-3000: $l = 2500$ , $c f = 97$ , $f = 28$ , $h = 500$
Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 2500 + \frac{3}{28} \times 500 = 2500 + 53.57 = 2553.57$
Mean (step deviation): class marks 1250, 1750, ..., 4750; take $a = 2750$ , $h = 500$ , $u_{i} = - 3, - 2, - 1, 0, 1, 2, 3, 4$
$f_{i} u_{i}$ : $- 72, - 80, - 33, 0, 30, 44, 48, 28$ ; $\sum f_{i} u_{i} = - 35$
Mean $= 2750 + \frac{- 35}{200} \times 500 = 2750 - 87.5 = 2662.5$
Median expenditure ≈ ₹2553.57, mean expenditure = ₹2662.50

Also asked in: 2023 Standard 30/4/2, 2023 Standard 30/4/3

Q355 marksLong AnswerStatistics CBSE 2023 · Standard 30/6/1

250 apples of a box were weighed and the distribution of masses of the apples is given in the following table :
Mass (in grams): 80 – 100, 100 – 120, 120 – 140, 140 – 160, 160 – 180
Number of apples: 20, 60, 70, $x$ , 60
(i) Find the value of $x$ and the mean mass of the apples. (3)
(ii) Find the modal mass of the apples. (2)

Show answer & solution

Answer: (i) x = 40, mean mass = 134.8 g (ii) modal mass = 125 g

(i) 20 + 60 + 70 + x + 60 = 250, so x = 40.
Class marks: 90, 110, 130, 150, 170.
$\sum f_{i} x_{i} = 1800 + 6600 + 9100 + 6000 + 10200 = 33700$
Mean $= \frac{33700}{250} = 134.8$ g
(ii) Modal class is 120 – 140 (frequency 70); $l = 120$ , $f_{1} = 70$ , $f_{0} = 60$ , $f_{2} = 40$ , $h = 20$ .
Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 120 + \frac{10}{40} \times 20 = 125$ g

Also asked in: 2023 Standard 30/6/2, 2023 Standard 30/6/3

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