Statistics Class 11 OP Malhotra Exe-28C ISC Maths Ch-28 Solutions. In this article you would learn about MODE. 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 Mathematics.

Statistics Class 11 OP Malhotra Exe-28C ISC Maths Solutions Ch-28

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-28	Statistics
Writer	O.P. Malhotra
Exe-28(C)	MODE

MODE

Statistics Class 11 OP Malhotra Exe-28C ISC Maths Ch-28 Solutions.

Que-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

Sol: (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

Que-2: Find the mode from the following data in question 2 to 7.

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

Que-3:

Sol: The table of values is given as under:

∴ Mean = Σfx/Σf = 652/41
Here, (N+1)/2 = (41+1)/2 = 21
Thus Md = 16
∴ Mode = 3 median – 2 mean = 3 × 16 – 2 × 652/41 ≃ 16.19

Que-4:

Sol: 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+{fm−(fm−1)} / [2fm−(fm−1)−(fm+1)] × i
= 10 + {(45−10)/(90−10−12)} × 10
= 10 + (350/68)
= 10 + 5.1470 = 15.147

Que-5:

Sol: 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+{fm−(fm−1)} / [2fm−(fm−1)−(fm+1)] × i
= 30 + {(60−28)/(120−28−38} × 10
= 30 + (320/56) = 35.71

Que-6:

Sol: Given class-intervals are discontinuous we make it continuous with the help of adjustment factor.
Here adjustment factor = (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+{fm−(fm−1)} / [2fm−(fm−1)−(fm+1)] × i
= 15.5 + {(32−16)/(64−16−24)} × 5
= 15.5 + (80/24) ≃ 18.84

Que-7:

Sol: 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+{fm−(fm−1)} / [2fm−(fm−1)−(fm+1)] × i
= 58 + {(25−0)/(50−8)} × 4 = 60.38

Que-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?

Sol: 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+{fm−(fm−1)} / [2fm−(fm−1)−(fm+1)] × i
48.6 = 40 + {(44−20)/(88−20−f)} × 15
⇒ 8.6 = 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.

Que-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)

Sol: (i) Required Mean = (17+32+35+33+15+21+41+32+11+18+20+22+11+15+35+23+38+12)/18
= 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)
Md = size of [(n/2)th + {(n/2)+1}th item]/2
= size of [(9+10)/2] = (21+22)/2 = 21.5
The observations 11, 15, 32 and 35 repeated equal no. of times i.e. 2 times. Thus, mode is undefined.

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

∴ Mean = Σfx/Σf = 1825/130 ≃ 14.04
Here, (N+1)/2 = (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.

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

Sol: 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+{fm−(fm−1)} / [2fm−(fm−1)−(fm+1)] × i
⇒ 240 = 200 + {(270−230)/(540−230−f)} × 100
⇒ 40 = 4000/(310−f) ⇒ 40(310 – f) = 4000
⇒ 12400 – 40f = 4000
⇒ 40f = 8400
⇒f = 210

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

Sol: The table of values is given as under:

Here (N+1)/2 = (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 {3( N+1)}/4 = {3×356}/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, {5( N+1)}/10 = {5×356}/10 = 178
∴ D6 = size of 178th item = 5

P65 ; Here, {65( N+1)}/100 = {65×356}/100 = 231.4
∴ P65 = size of 231.4th item = 6

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

Sol: The table of values is given as under:

Then by Step-deviation method, we have
Mean x¯ = A + (Σfd′/N) × i = 45 – (15/393) × 10 = 44.62
Here N/2 = 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 + [{(N/2)−C}/f] × i = 40 + {(196.5−148)/78} × 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
Thus, Mode = l+{fm−(fm−1)} / [2fm−(fm−1)−(fm+1)] × i
= 50 + {(80−78)/(160−78−70)} × 10 = 50 + (20/12) = 51.67
P10: Here (10×N)/100 = (10×393)/100 = 39.3 which lies in 10 – 20
Here l = 10; c = 15; f = 25; i = 10
P90 = l + [{(90N/100)−C}/f] × i = 60 + {(353.7−306)/70} × 10 = 66.81
The required limits between which 80% students have secured marks be 19.72 and 66.81.

–: End of Statistics Class 11 OP Malhotra Exe-28C ISC Maths Ch-28 Solutions. :–

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