OP Malhotra Class 11 Maths Solutions Chapter 28 Statistics Ex 28(c)

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.

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

Question 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 M_d = 16
∴ Mode = 3 median – 2 mean = 3 × 16 – 2 × \(\frac{652}{41}\) ≃ 16.19

Question 4.

Solution:
From given data, maximum frequency be 45 and thus 10 – 20 be the modal class.
Here l = 10; f_m = 45 ; f_m-1 = 10; f_m+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.

Solution:
Here maximum frequency be 60 and modal class be 30 – 40.
Here, l = 30; f_m = 60; f_m-1 = 28 ; f_m+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.

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; f_m = 32 ; f_m-1 = 16; f_m+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.

Solution:
The table of values is given as under:

Here Modal class be 58 – 62.
Here
l = 58; f_m = 25 ; f_m+1 = 8 ; i = 4 ; f_m-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.

(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; f_m = 44 ; f_m-1 = 20 ; f_m+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 f_m, f_m-1 and f_m+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)

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 :

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

Here l = 200; f_m = 270 ; f_m-1 = 230 ; f_m+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 :

Solution:
The table of values is given as under:

Here \(\frac{N+1}{2}\) = \(\frac{355+1}{2}\) = 178 ∴ M_d = 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.

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
∴ M_d = \(l+\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{f} \times i\) = 40 + \(\frac{196.5-148}{78} \times 10\) = 40 + 6.2195 = 46.22
Clearly modal class be 50 – 60.
Here l = 50 ; f_m = 80 ; f_m-1 = 78 ; f_m+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.