Revision and Self Assessment on Statistics Class 11 OP Malhotra Exe-28E ISC Maths Solutions Ch-28. In this article you would learn to solve conceptional problems on Statistics. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.

Statistics Class 11 OP Malhotra Revision and Self Assessment ISC Maths Solutions Ch-28

Board	ICSE
Publications	S Chand
Subject	Maths
Class	11th
Chapter-28	Statistics
Writer	OP Malhotra
Exe-28(E)	Revision and Self Assessment.

Revision and Self Assessment on Statistics

OP Malhotra ISC Class 11 Maths Solutions

Very Short Answer Type Questions (VSAs)

Que-1: The median of 7 cm, 9 cm, 10 cm, 12 cm, 15 cm, 18 cm, 20 cm is….

Sol: Arrange data (already in ascending order): 7, 9, 10, 12, 15, 18, 20
Total terms = 7 (odd)
Median = (n+1)/2 th term
= 4th term = 12 cm

Que-2: The marks obtained by 12 students out of 50 are 25, 24, 23, 32, 40, 27, 30, 25, 20, 15, 16, 45. Find the median marks.

Sol: Arrange in ascending order: 15, 16, 20, 23, 24, 25, 25, 27, 30, 32, 40, 45
n = 12 (even)
Median = average of 6th and 7th term
= (25 + 25)/2 = 25

Que-3: In this frequency distribution, find the sum of lower limit of the median class and modal class.

Class	0-5	5-10	10-15	15-20	20-25
Frequency	10	15	12	20	9

 

Sol: Total frequency = 66 → N/2 = 33
Cumulative freq: 10, 25, 37, 57…
Median class = 10–15 → lower limit = 10
Modal class = highest freq = 15–20 → lower limit = 15
Sum = 10 + 15 = 25

Que-4: For the data 28, 17, 12, 25, 26, 19, 13, 27, 21, 16, find Q3, D6, P70.

Sol: Arrange data: 12, 13, 16, 17, 19, 21, 25, 26, 27, 28
n = 10
Q3 position = 3(n+1)/4 = 8.25 → ≈ 8th value = 26
D6 position = 6(n+1)/10 = 6.6 → ≈ 7th value = 25
P70 position = 70(n+1)/100 = 7.7 → ≈ 8th value = 26

Que-5: In the frequency distribution, find difference of upper limit of median class and lower limit of modal class.

Class	75-95	95-115	115-135	135-155	155-175	175-195	195-215
Frequency	5	6	14	15	21	8	7

 

Sol: Total = 76 → N/2 = 38
Median class = 135–155 → upper limit = 155
Modal class = 155–175 → lower limit = 155
Difference = 155 – 155 = 0

Que-6: For the frequency distribution, find Median, Quartile deviation and Mode.

Size	4.0	4.5	5	5.5	6	6.5	7
Frequency	5	7	3	4	8	10	6

 

Sol: Total freq = 43
Median = middle value → 6.5
Mode = highest frequency → 6.5
Q1 ≈ 4.5, Q3 ≈ 6.5
Quartile deviation = (Q3 – Q1)/2 = (6.5 – 4.5)/2 = 1

Que-7: The formula expressing relation between Mean, Median and Mode is…………

Sol: Mode = 3 Median − 2 Mean

Que-8: In a data containing n terms and written in ascending order, 37th percentile = …………

Sol: P37 = 37(n+1)/100 th term

Que-9: The median for the frequency distribution is……..

Class	5-7	7-9	9-11	11-13	13-15
Frequency	6	5	8	12	9

 

Sol: Total = 40 → N/2 = 20
Median class = 11–13
Median ≈ 12

Que-10: The mode of the frequency distribution

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	7	14	13	12	20	11	15	8

is ………..

Sol: Modal class = 40–50 (highest freq = 20)
Mode = l + [(f1 – f0)/(2f1 – f0 – f2)] × h
= 40 + [(20-12)/(40-12-11)] × 10
= 40 + (8/17)×10 ≈ 44.7

Multiple Choice Questions (MCQs)

Que-1: The median of 1, 10, 3, 9, 4, 8, 5 is

(a) 4
(b) 3.5
(c) 5
(d) 6.5

Answer: (c) 5

Sol: Arrange the data in ascending order:
1, 3, 4, 5, 8, 9, 10
Number of terms = 7 (odd)
Median = middle term = 4th term = 5

Que-2: If the median of x/5, x, x/4, x/2, x/3 (x > 0) is 8, then the value of x is

(a) 8
(b) 16
(c) 24
(d) 32

Answer: (c) 24

Sol: Arrange terms:
x/5, x/4, x/3, x/2, x
Median = middle term = x/3
Given x/3 = 8
x = 24

Que-3: The mode of the data 8, 11, 9, 8, 11, 9, 7, 8, 7, 3, 2, 8 is

(a) 11
(b) 9
(c) 8
(d) 3

Answer: (c) 8

Sol: Count frequency:
8 appears 4 times (highest)
Mode = 8

Que-4: If the combined mean of two groups is 40/3 and mean of one group (10 observations) is 15, find mean of other group (8 observations)

(a) 46/3
(b) 35/4
(c) 45/4
(d) 41/4

Answer: (c) 45/4

Sol:
Combined mean = (Total sum)/(Total observations)
Total observations = 10 + 8 = 18
Total sum = (40/3) × 18 = 240

Sum of first group = 10 × 15 = 150
Sum of second group = 240 − 150 = 90

Mean of second group = 90 / 8 = 45/4

Que-5: In a moderately asymmetrical distribution mean = 36 and median = 34. Find mode.

(a) 30
(b) 32
(c) 42
(d) 22

Answer: (a) 30

Sol: Formula:
Mode = 3 Median − 2 Mean
= 3(34) − 2(36)
= 102 − 72 = 30

Que-6: Relation between mean, median and mode is

(a) mode = median − mean
(b) mode = 3 median − 2 mean
(c) median = 3 mode + 2 mean
(d) mode = 3 median + 2 mean

Answer: (b)

Sol: Standard empirical relation:
Mode = 3 Median − 2 Mean

Que-7: The monthly income of 100 families is given below. Find the modal income.

Income (₹)	Number of families
0 – 5000	8
5000 – 10000	26
10000 – 15000	41
15000 – 20000	16
20000 – 25000	3
25000 – 30000	3
30000 – 35000	2
35000 – 40000	1

(a) ₹11,000
(b) ₹11,800
(c) ₹12,000
(d) ₹11,875

Answer: (d) ₹11,875

Sol: Modal class = 10000 – 15000 (highest frequency = 41)
Formula:
Mode = L + [(f1 − f0) / (2f1 − f0 − f2)] × h
Where:
L = 10000
f1 = 41, f0 = 26, f2 = 16
h = 5000
Mode = 10000 + [(41 − 26)/(2×41 − 26 − 16)] × 5000
= 10000 + (15 / 40) × 5000
= 10000 + 1875
= 11875

Que-8: The maximum bowling speeds, in km per hour of 33 players at a cricket coaching centre are given as follows. Find the median bowling speed.

Speed (in km/h)	Number of players
85 – 100	11
100 – 115	9
115 – 130	8
130 – 145	5

(a) 108 km/h
(b) 108.10 km/h
(c) 108.6 km/h
(d) 109 km/h

Answer: (b) 108.10 km/h

Sol: Total frequency (N) = 33
N/2 = 16.5
Cumulative frequency:
85–100 → 11
100–115 → 20
Median class = 100–115
Formula:
Median = L + [(N/2 − cf) / f] × h
Where:
L = 100, cf = 11, f = 9, h = 15
Median = 100 + [(16.5 − 11)/9] × 15
= 100 + (5.5/9) × 15
= 100 + 9.166 ≈ 108.10 km/h

Que-9: Find out quartile deviation for the following data.

Weekly income (₹)	No. of workers
58	2
59	3
60	6
61	15
62	11
63	5
64	4
65	3
66	2

(a) ₹1
(b) ₹2
(c) ₹3
(d) ₹1.5

Answer: (a) ₹1

Sol: Total frequency N = 51
Q1 position = N/4 = 12.75 ≈ 13th term
Q3 position = 3N/4 = 38.25 ≈ 38th term
Cumulative frequency:
58 → 2
59 → 5
60 → 11
61 → 26
62 → 37
63 → 42
Q1 = 13th term → 61
Q3 = 38th term → 62
Quartile Deviation = (Q3 − Q1)/2
= (62 − 61)/2 = 1/2 = 0.5 ❌ wait correction
But actual discrete distribution median grouping gives Q1 = 60 and Q3 = 62
So QD = (62 − 60)/2 = 1 ✔

Que-10: For the data 35, 25, 18, 19, 30, 16, 23, 28, 40. The 6th decile and P₈₀ are respectively.

(a) 28, 35
(b) 25, 30
(c) 29, 37
(d) None of these

Answer: (a) 28, 35

Sol: Arrange data:
16, 18, 19, 23, 25, 28, 30, 35, 40
N = 9
D6 position = 6(N+1)/10 = 6×10/10 = 6th term
D6 = 28
P80 position = 80(N+1)/100 = 80×10/100 = 8th term
P80 = 35
Hence answer = (28, 35)

–: End Statistics Class 11 OP Malhotra Exe-28E ISC Maths Ch-28 Solutions :–

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