Continuous practice using ISC Class 11 Maths Solutions S Chand Chapter 28 Statistics Ex 28(c) can lead to a stronger grasp of mathematical concepts.

## S Chand Class 11 ICSE Maths Solutions Chapter 28 Statistics Ex 28(c)

Question 1.
Find the mode of the following data :
(i) 3, 4, 7, 11, 4, 3, 4, 5, 6, 4, 1, 2, 4, 4
(ii) Size of shoes : 4, 4.5, 5, 4.5, 5.5, 5, 6, 4.5, 4, 4.5
(iii) Wages (₹) : 100, 120, 100, 120, 130, 120, 120, 130, 120, 100
(iv) Runs in an innings : 18, 32, 0, 40, 60, 69, 33, 69, 35, 11, 20
Solution:
(i) Arranging the given data in ascending order; we have
1, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 6, 7, 11
Here 4 repeated maximum no. of times.
∴ required mode = 4

(ii) Arranging the given data in ascending order : 4, 4, 4.5, 4.5, 4.5, 4.5, 5, 5, 5.5, 6
Here 4.5 repeated maximum no. of times.
∴ required mode = 4.5

(iii) Arranging the given data in ascending order :
100, 100, 100, 120, 120, 120, 120, 120, 130, 130
Here 120 repeated maximum no. of times i.e. 5 times. Thus required mode = 120

(iv) Arranging the given data in ascending order: 0, 11, 18, 20, 32, 33, 35, 40, 60, 69, 69
Here 69 repeated maximum no. of times
∴ required mode = 69

Question 2.
Find the mode from the following data in question 2 to 7.

 Marks 10 12 15 22 25 35 45 50 60 Number of students 4 6 10 14 20 19 10 6 3

Solution:
Here maximum frequency be 20 and corresponding observation be 25. Thus required modal marks be 25.

Question 3.

 Size 8 10 12 14 16 18 20 22 24 Frequency 3 5 1 7 8 6 4 2 3

Solution:
The table of values is given as under:

∴ Mean = $$\frac{\Sigma f x}{\Sigma f}$$ = $$\frac{652}{41}$$
Here, $$\frac{N+1}{2}$$ = $$\frac{41+1}{2}$$ = 21
Thus Md = 16
∴ Mode = 3 median – 2 mean = 3 × 16 – 2 × $$\frac{652}{41}$$ ≃ 16.19

Question 4.

 Class interval 0-10 10-20 20-30 30-40 Frequency 10 45 12 3

Solution:
From given data, maximum frequency be 45 and thus 10 – 20 be the modal class.
Here l = 10; fm = 45 ; fm-1 = 10; fm+1 = 12 ; i = 10
Mode = $$l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i$$ = 10 + $$\frac{45-10}{90-10-12} \times 10$$ = 10 + $$\frac{350}{68}$$ = 10 + 5.1470 = 15.147

Question 5.

 Marks 0-10 10-20 20-30 30-40 40-50 Number of students 8 16 26 60 38

Solution:
Here maximum frequency be 60 and modal class be 30 – 40.
Here, l = 30; fm = 60; fm-1 = 28 ; fm+1 = 38 ; i = 10
∴ Mode = $$l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i$$ = 30 + $$\frac{60-28}{120-28-38} \times 10$$ =30 + $$\frac{320}{56}$$ = 35.71

Question 6.

 Marks Frequency Marks Frequency 1-5 7 26-30 18 6-10 10 31-35 10 11-15 16 36-40 5 16-20 32 41-45 1 21-25 24

Solution:
Given class-intervals are discontinuous we make it continuous with the help of adjustment factor.
Here adjustment factor = $$\frac{6-5}{2}$$ = 0.5
We subtract 0.5 from each lower limit and add 0.5 to each upper limit.

Here modal class be 15.5 – 20.5
∴ l = 15.5; fm = 32 ; fm-1 = 16; fm+1 = 24; i = 5
Thus,
Mode = $$l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i$$
= 15.5 + $$\frac{32-16}{64-16-24} \times 5$$
= 15.5 + $$\frac{80}{24}$$ ≃ 18.84

Question 7.

 Mid-value 60 64 68 72 76 Frequency 25 8 10 6 4

Solution:
The table of values is given as under:

 Mid-value Class Intervals Frequency(f) 60 58-62 25 64 62-66 8 68 66-70 10 72 70-74 6 76 74-78 4 Σf = N = 53

Here Modal class be 58 – 62.
Here
l = 58; fm = 25 ; fm+1 = 8 ; i = 4 ; fm-1 = 0
Thus Mode = $$l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i$$ = 58 + $$\frac{25-0}{50-8}$$ × 4 = 60.38

Question 8.
(i) The mode of the following frequency distribution is 48.6 Find the unknown frequency.

 Class 20-25 25-40 40-55 55-70 70-85 Frequency 6 20 44 ? 3

(ii) If the frequency of the class 70-85 is 13 instead of 3, then what difference will it make?
Solution:
Let the missing frequency be f.
Here modal class be 40 – 55.
∴ l = 40; fm = 44 ; fm-1 = 20 ; fm+1 = f; i = 15
Thus,
Mode = $$l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m-1}} \times i$$
48.6 = 40 + $$\frac{44-20}{88-20-f} \times 15$$
⇒ 8.6 = $$\frac{360}{68-f}$$
⇒ 584.8 – 8.6f = 360
⇒ 8.6(68 – f) = 360
⇒ 584.8 – 8.6f = 360
⇒ 8.6 f = 224.8
⇒ f = 26.14
If the frequency of class 70 – 85 is 13 instead of 3.
Then value of mode is unaffected as value of mode affected if there are changes in the values of fm, fm-1 and fm+1.

Question 9.
Find the mean, median and mode of the following :
(i) The data 17, 32, 35, 33, 15, 21, 41, 32, 11, 18, 20, 22, 11, 15, 35, 23, 38, 12.
(ii)

 Size of item 11 12 13 14 15 16 17 Frequency 8 13 25 37 23 15 9

Solution:
(i) Required Mean = $$\frac{17+32+35+33+15+21+41+32+11+18+20+22+11+15+35+23+38+12}{18}$$
= $$\frac{431}{18}$$ = 23.94
Arranging the given data in ascending order ; we have
11, 11, 12, 15, 15, 17, 18, 20, 21, 22, 23, 32, 32, 33, 35, 35, 38, 41
Here no. of observations = n = 18(even)

The observations 11, 15, 32 and 35 repeated equal no. of times i.e. 2 times. Thus, mode is illdefined.

(ii) The table of values is given as under:

∴ Mean = $$\frac{\Sigma f x}{\Sigma f}$$ = $$\frac{1825}{130}$$ ≃ 14.04
Here, $$\frac{N+1}{2}$$ = $$\frac{130+1}{2}$$ = 65.5 ∴ Median = 14
Further maximum frequency be 37 and corresponding value of x i.e. 14 gives the required value of mode.

Question 10.
The mode of the following distribution is 240. Find out the missing frequency :

 Size Frequency Size Frequency 0-100 140 300-400 ? 100-200 230 400-500 150 200-300 270 500-600 140

Solution:
Let the missing frequency be f.
Clearly the maximum frequency be 270 and modal class be 200 – 300.

Here l = 200; fm = 270 ; fm-1 = 230 ; fm+1 = f ; i = 100
Thus, Mode = $$l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i$$ ⇒ 240 = 200 + $$\frac{(270-230)}{540-230-f} \times 100$$
⇒ 40 = $$\frac{4000}{310-f}$$ ⇒ 40(310 – f) = 4000 ⇒ 12400 – 40f = 4000 ⇒ 40f = 8400 ⇒f = 210

Question 11.
Find the median, mode, third quartile, 5th decile and 65th percentile from the following :

 Value of the item 1 2 3 4 5 6 7 8 9 10 Frequency 25 30 38 45 46 58 32 34 25 22

Solution:
The table of values is given as under:

Here $$\frac{N+1}{2}$$ = $$\frac{355+1}{2}$$ = 178 ∴ Md = 5
Here maximum Frequency be 58 and corresponding value of x i.e. 6 be the required value of mode.

Q3 : Here $$\frac{3(\mathrm{~N}+1)}{4}$$ = $$\frac{3 \times 356}{4}$$ = 267 and corresponding c.f be 274 and the corresponding value of x gives the required value of Q3. Thus Q3 = 7

D5 : Here, $$\frac{5(\mathrm{~N}+1)}{10}$$ = $$\frac{5 \times 356}{10}$$ = 178
∴ D6 = size of 178th item = 5

P65 ; Here, $$\frac{65(\mathrm{~N}+1)}{100}$$ = $$\frac{65 \times 356}{100}$$ = 231.4
∴ P65 = size of 231.4th item = 6

Question 12.
Find median, mean and modal marks and also determine the limits between which 80% of the students have secured marks.

 Marks No. of students Marks No. of students 0-10 15 50-60 80 10-20 25 60-70 70 20-30 52 70-80 10 30-40 56 80-90 5 40-50 78 90-100 2

Solution:
The table of values is given as under:

Then by Step-deviation method, we have
Mean $$\bar{x}$$ = A + $$\frac{\Sigma f d^{\prime}}{\mathrm{N}}$$ × i = 45 – $$\frac { 15 }{ 393 }$$ × 10 = 44.62
Here $$\frac { N }{ 2 }$$ = $$\frac { 393 }{ 2 }$$ = 196.5 which lies in 40 – 50.
Thus median class be 40 – 50.
Here
l = 40; f = 78; c = 148 ; i = 10
∴ Md = $$l+\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{f} \times i$$ = 40 + $$\frac{196.5-148}{78} \times 10$$ = 40 + 6.2195 = 46.22
Clearly modal class be 50 – 60.
Here l = 50 ; fm = 80 ; fm-1 = 78 ; fm+1 = 70 ; i = 10
∴ Mode = $$l+\frac{f_m-f_{m-1}}{2 f_m-f_{m-1}-f_{m+1}} \times i$$ = 50 + $$\frac{80-78}{160-78-70}$$ × 10 = 50 + $$\frac { 20 }{ 12 }$$ = 51.67
P10: Here $$\frac{10 \times N}{100}$$ = $$\frac{10 \times 393}{100}$$ = 39.3 which lies in 10 – 20
Here l = 10; c = 15; f = 25; i = 10
$$\mathrm{P}_{90}=l+\frac{\frac{90 \mathrm{~N}}{100}-\mathrm{C}}{f} \times i$$ = $$60+\frac{353.7-306}{70} \times 10$$ = 66.81
The required limits between which 80% students have secured marks be 19.72 and 66.81.