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Maths 4 Mark Question

Measures Of Central Tendency · Maharashtra Board STD 9 (MSBSHSE - English Medium). 22 questions.

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Question 14 Marks
Duration of sunshine(in hours) in Amritsar for first 10 days of August 1997 as reported by the Meterological Department are given as follows: 9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, 10.9
  1. Find the mean $\overline{\text{X}}$
  2. Verify that $\sum\limits_{\text{i}=1}^{10}\Big(\text{X}-\overline{\text{X}}\Big)=0$
Answer
Duration of sunshine (in hours) for 10 days are = 9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, 10.9
  1. $\text{Mean X}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$
$=\frac{9.6+5.2+3.5+1.5+1.6+2.4+2.6+8.4+10.3+10.9}{10}$

$=\frac{56}{10}=5.6$
  1. $\text{L.H.S.}=\sum\limits_{\text{i}=1}^{10}\Big(\text{x}^\text{i}-\overline{\text{X}}\Big)$
$=\Big(\text{X}_1-\overline{\text{X}}\Big)+\Big(\text{X}_2-\overline{\text{X}}\Big)+\Big(\text{X}_3-\overline{\text{X}}\Big)+\dots+\Big(\text{X}_{10}-\overline{\text{X}}\Big)$

$= (9.6-5.6) + (5.2 - 5.6) + (3.5 - 5.6) + (1.5 - 5.6) + (1.6 - 5.6) \\+ (2.4 - 5.6) + (2.6 - 5.6) + (8.4 - 5.6) + (10.3 - 5.6) + (10.9 - 5.6)$

$= 4 - 0.4 - 2.1 - 4.1 - 4 - 3.2 - 3 + 2.8 + 4.7 + 5.3$

$= 16.8-16.8 = 0$
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Question 24 Marks
If the mean of the following data is 15, find p.
f:
6
p
6
10
5
x:
5
10
15
20
25
Answer
x
f
fx
5
6
30
10
p
10p
15
6
90
20
10
200
25
5
125
 
N = p + 27
$\sum\text{fx}=10\text{p}+445$
It is given that,
Mean = 15
$\Rightarrow\frac{\sum\text{fx}}{\text{N}}=15$
$\Rightarrow\text{10p}+\text{445p}+27=15$
$\Rightarrow\text{10p}+445=15\times(\text{p}+27)$
$\Rightarrow\text{10p}+445=\text{15p}+405$
$\Rightarrow\text{15p}-\text{10p}=445-405$
$\Rightarrow\text{5p}=40$
$\Rightarrow\text{p}=405=8$
$\Rightarrow\text{p}=8$
$\therefore\text{p}=8$
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Question 34 Marks
Find the value of p, if the mean of the following distribution is 20.
x:
15
17
19
20 + p
23
f:
2
3
4
5p
6
Answer
x
f
fx
15
2
30
17
3
51
19
4
76
20+p
5p
$100p + 5p^2$
23
6
138
 
N = 5p + 15
$fx = 5p^2 + 100p + 295$
It is given that,
Mean = 20
$\Rightarrow\frac{\sum\text{fx}}{\text{N}}=20$
$\Rightarrow\frac{\text{5p}^2+\text{100p}+295}{\text{5p}+15}=20$
$\Rightarrow 5p^2 + 100p + 295 = 20(5p + 15)$
$\Rightarrow 5p^2 + 100p + 295 = 100p + 300$
$\Rightarrow 5p^2 = 300 − 295$
$\Rightarrow 5p^2 = 5$
$\Rightarrow p^2 = 1$
$\Rightarrow\text{p}=\pm1$
Frequency can’t be negative.
Hence, value of p is $1$
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Question 44 Marks
Find the mean of the following distribution:
x:
10
12
20
25
35
f:
3
10
15
7
5
Answer
x
f
fx
10
3
30
12
10
120
20
15
300
25
7
175
35
5
175
 
N = 40
$\sum\text{fx}=800$
$\therefore\text{Mean }\bar{\text{x}}=\frac{\sum\text{fx}}{\text{N}}$
$=\frac{800}{40}=20$
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Question 54 Marks
Find the missing frequency (p) for the following distribution whose mean is 7.68
x:
3
5
7
9
11
13
f:
6
8
15
p
8
4
Answer
x
f
fx
3
6
18
5
8
40
7
15
105
9
P
9p
11
8
88
13
4
52
 
N = p + 41
$\sum\text{fx}=\text{9p}+303$
It is given that,
Mean = 7.68
$\Rightarrow\frac{\sum\text{fx}}{\text{N}}=7.68$
⇒ 9p + 303p + 41 = 7.68
⇒ 9p + 303 = 7.68p + 314.88
⇒ 9p − 7.68p = 314.88 - 303
⇒ 1.32p = 11.88
⇒ p = 11.881.32 = 9
⇒ p = 9
$\therefore$ P = 9
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Question 64 Marks
The mean of the following data is 20.6 Find the value of p.
x:
10
15
p
25
35
f:
3
10
25
7
5
Answer
x
f
fx
10
3
30
15
10
150
P
25
25p
25
7
175
35
5
175
 
N = 50
$\sum\text{fx}=25\text{p}+530$
It is given that,
Mean = 20.6
$\Rightarrow\frac{\text{25p}+530}{50}=20.6$
$\Rightarrow\text{25p}+530=20.6\times50$
$\Rightarrow\text{25p}=1030-530$
$\Rightarrow\text{25p}=500$
$\Rightarrow\text{p}=\frac{500}{25}=20$
$\Rightarrow\text{p}=20$
$\therefore\text{p}=20$
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Question 74 Marks
Candidates of four schools appear in a mathematics test. The data were as follows:
Schools
No. of Candidates
Average Score
I
60
75
II
48
80
III
Not Available
55
IV
40
50
If the average score of the candidates of all four schools is 66, Find the number of candidates that appeared from school III.
Answer
Schools
No. of Candidates
Average Score
I
60
75
II
48
80
III
x
55
IV
40
50
Given the average score of all schools = 66
$\Rightarrow\frac{\text{N}_1\overline{\text{X}}_1+\text{N}_2\overline{\text{X}}_2+\text{N}_3\overline{\text{X}}_3+\text{N}_4\overline{\text{X}}_4}{\text{N}_1+\text{N}_2+\text{N}_3+\text{N}_4}=66$
$\Rightarrow\frac{60\times75+48\times80+\text{x}\times55+40\times50}{60+48+\text{x}+40}=66$
$\Rightarrow\frac{4500+3840+\text{55x}+2000}{148+\text{x}}=66$
$\Rightarrow\frac{10340+\text{55x}}{148+\text{x}}=66$
$\Rightarrow10340+55\text{x}=66\text{x}+9768$
$\Rightarrow10340-9768=\text{66x}-\text{55x}$
$\Rightarrow\text{11x}=572$
$\Rightarrow\text{x}=\frac{572}{11}=52$
$\therefore$ No. of candidates appeared from school III = 52
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Question 84 Marks
Find the values of n and $\overline{\text{X}}$ in the following case:$\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-10)=30$ and $\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-6)=150$
Answer
Given $\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-10)=30$$\Rightarrow(\text{x}_1-10)+(\text{x}_2-10)+\dots+(\text{x}_\text{n}-10)=30$
$\Rightarrow(\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5+\dots+\text{x}_\text{n})\\-(10+10+10+10+\dots+10)=30$
$\Rightarrow\sum\text{x}-\text{10n}=30\dots(1)$
And $\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-6)=150$
$\Rightarrow(\text{x}_1-6)+(\text{x}_2-6)+\dots+(\text{x}_\text{n}-6)=150$
$\Rightarrow(\text{x}_1+\text{x}_2+\dots+\text{x}_\text{n})-(6+6+6+\dots+6)=150$
$\Rightarrow\sum\text{x}-\text{6n}=150\dots(2)$
By subtracting equation (1) from equation (2), we get
$\sum\text{x}-\text{6n}-\sum\text{x}+\text{10n}=150-30$
$\Rightarrow\text{4n}=120$
$\Rightarrow\text{n}=\frac{120}{4}=30$
Put value of n in equation (1)
$\sum\text{x}-10\times30=30$
$\Rightarrow\sum\text{x}-300=30$
$\Rightarrow\sum\text{x}=30+300=330$
$\therefore\overline{\text{x}}=\frac{\sum\text{x}}{\text{n}}=\frac{330}{30}=11$
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Question 94 Marks
Find the median of the following data: 41, 43, 127, 99, 61, 92, 71, 58, 57. If 58 is replaced by 85, what will be the new median?
Answer
Given the numbers are 41, 43, 127, 99, 61, 92, 71, 58, 57 Arranging the numbers in ascending order 41, 43, 57, 58, 61, 71, 92, 99, 127 n = 9 (odd)$\therefore\text{New Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$
$=\Big(\frac{\text{9}+1}{2}\Big)^{\text{th}}\text{value}$
If 58 is replaced by 85 Then the new values be in order are: 41, 43, 57, 61, 71, 85, 92, 99, 127$\therefore\text{New Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$
$=\Big(\frac{\text{9}+1}{2}\Big)^{\text{th}}\text{value}$
$=5^{\text{th}}\text{value}=71$
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Question 104 Marks
If the ratio of mode and median of a certain data is 6 : 5, then find the ratio of its mean and median.
Answer
Given that the ratio of mode and median of a certain data is 6 : 5. That is, MODE : MEDIAN = 6 : 5$\Rightarrow\frac{\text{MODE}}{\text{MEDIAN}}=\frac{6}{5}$
$\Rightarrow5\times\text{MODE}=6\times\text{MEDIAN}$
$\Rightarrow\text{MODE}=\frac{6}{5}\text{MEDIAN}$
We know that,$\text{MODE}=3\times\text{MEDIAN}-2\times\text{MEAN}$
$\Rightarrow2\times\text{MEAN}=\Big(3-\frac{6}{5}\Big)\text{MEDIAN}$
$\Rightarrow2\times\text{MEAN}=\frac{9}{5}\text{MEDIAN}$
$\Rightarrow\text{MEAN}=\frac{9}{10}\text{MEDIAN}$
$\Rightarrow\frac{\text{MEAN}}{\text{MEDIAN}}=\frac{9}{10}$
$\Rightarrow\text{MEAN}:\text{MEDIAN}=9:10$
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Question 114 Marks
Find the value of p for the following distribution whose mean is 16.6
x:
8
12
15
p
20
25
30
f:
12
16
20
24
16
8
4
Answer
x
f
fx
8
12
96
12
16
192
15
20
300
P
24
24p
20
16
320
25
8
200
30
4
120
 
N = 100
$\sum\text{fx} = 24\text{p} + 1228$
It is given that,
Mean = 16.6
$\Rightarrow\frac{\sum\text{fx}}{\text{N}}=16.6$
$\Rightarrow 24\text{p} + 1228 = 1660$
$\Rightarrow 24\text{p} = 1660 − 1228$
$\Rightarrow 24\text{p} = 432$
$\Rightarrow\text{p}=\frac{432}{24}=18$
$\Rightarrow \text{p} = 18$
$\therefore\text{p} = 18$
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Question 124 Marks
Five coins were simultaneously tossed 1000 times and at each, toss the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.
No. of heads per toss
No. of tosses
0
38
1
144
2
342
3
287
4
164
5
25
Total
1000
Answer
No. of heads per toss (x)
No. of tosses (f)
fx
0
38
0
1
144
144
2
342
684
3
287
861
4
164
656
5
25
125
 
N = 1000
$\sum\text{fx}=2470$
$\therefore$ Mean number of heads per toss $=\frac{\sum\text{fx}}{\text{N}}$
$=\frac{2470}{1000}$
$=2.47$
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Question 134 Marks
Calculate the mean for the following distribution:
x:
5
6
7
8
9
f:
4
8
14
11
3
Answer
X
f
fx
5
4
20
6
8
48
7
14
98
8
11
88
9
3
27
 
N = 40
$\sum\text{fx}=281$
$\therefore\text{Mean}\ \bar{\text{x}}=\frac{\sum\text{fx}}{\text{N}}$
$=\frac{281}{40}=7.025$
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Question 144 Marks
Find the mean of the following data:
x:
19
21
23
25
27
29
31
f:
13
15
16
18
16
15
13
Answer
x
f
fx
19
13
247
21
15
315
23
16
368
25
18
450
27
16
432
29
15
435
31
13
403
 
N = 106
$\sum\text{fx}=2650$
$\therefore\text{Mean }\bar{\text{x}}=\frac{\sum\text{fx}}{\text{N}}$
$=\frac{2650}{106}=25$
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Question 154 Marks
Find the missing value of p for the following distribution whose mean is 12.58.
x:
5
8
10
12
p
20
25
f:
2
5
8
22
7
4
2
Answer
x
f
fx
5
2
10
8
5
40
10
8
80
12
22
264
P
7
7p
20
4
80
25
2
50
 
N = 50
$\sum\text{fx}=\text{7p}+524$
It is given that,
Mean = 12.58
$\Rightarrow\frac{\sum\text{fx}}{\text{N}}=12.38$
$\Rightarrow\frac{\text{7p}+524}{50}12.58$
⇒ 7p + 524 = 629
⇒ 7p = 629 - 524
⇒ 7p = 105
⇒ p = 1057 = 15
⇒ p = 15
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Question 164 Marks
Find the values of n and $\overline{\text{X}}$ in the following case:$\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-12)=-10$ and $\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-3)=62$
Answer
Given $\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-12)=-10$$\Rightarrow(\text{x}_1-12)+(\text{x}_2-12)+\dots+(\text{x}_\text{n}-12)=-10$
$\Rightarrow(\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5+\dots+\text{x}_\text{n})\\-(12+12+12+12+\dots+12)=-10$
$\Rightarrow\sum\text{x}-\text{12n}=-10\dots(1)$
And $\sum\limits^\text{n}_{\text{i}=1}(\text{x}_\text{i}-3)=62$
$\Rightarrow(\text{x}_1-3)+(\text{x}_2-3)+\dots+(\text{x}_\text{n}-3)=62$
$\Rightarrow(\text{x}_1+\text{x}_2+\dots+\text{x}_\text{n})-(3+3+3+\dots+3)=62$
$\Rightarrow\sum\text{x}-\text{3n}=62\dots(2)$
By subtracting equation (1) from equation (2), we get
$\sum\text{x}-\text{3n}-\sum\text{x}+\text{12n}=62+10$
$\Rightarrow\text{9n}=72$
$\Rightarrow\text{n}=\frac{72}{9}=8$
Put value of n in equation (1)
$\sum\text{x}-12\times8=-10$
$\Rightarrow\sum\text{x}-96=-10$
$\Rightarrow\sum\text{x}=96-10=86$
$\therefore\overline{\text{x}}=\frac{\sum\text{x}}{\text{n}}=\frac{86}{8}=10.75$
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Question 174 Marks
Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50
x:
10
30
50
70
90
f:
17
$f_1$
32
$f_2$
19
Answer
x
f
fx
10
$17$
$170$
30
 $f _1$ $30 f _1$
50
$32$
$1600$
70
$f_2$
$70f_2$
90
$19$
$1710$
 
$N = 120$
$\sum\text{fx}=3480+30\text{f}_1+70\text{f}_2$

It is given that
Mean $=50$
$\Rightarrow \frac{\sum fi }{ N }=50$
$\Rightarrow \frac{3480+30 f _1+70 f _2}{N}=50$
$\Rightarrow 3480+30 f _1+70 f _2=50 \times 120$
$\Rightarrow 30 f _1+70 f _2=6000-3480$
$\Rightarrow 10\left(3 f _1+7 f _2\right)=10(252)$
$\Rightarrow 3 f _1+7 f _2=252 \ldots$ (1) $[\because$ Divide by 10$]$
And $N=20$
$\Rightarrow 17+ f _1+32+ f _2+19=120$
$\Rightarrow 68+ f _1+ f _2=120$
$\Rightarrow f _1+ f _2=120-68$
$\Rightarrow f _1+ f _2=52$
Multiply with 3 on both sides
$\Rightarrow 3 f_1+3 f_2=156 \ldots \text { (2) }$
Subtracting equation (2) from equation (1)
$\Rightarrow 3 f_1+7 f_2-3 f_1-3 f_2=252-156$
$\Rightarrow 4 f_2=96$
$\Rightarrow f_2=\frac{96}{4}=24$
Put the value of $f_2$ in equation (1)
$\Rightarrow 3 f_1+7 \times 24=252$
$\Rightarrow 3 f_1=252-168$
$\Rightarrow f_1=\frac{84}{3}=28$
$\Rightarrow f_1=28$
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Question 184 Marks
The weights (in kg) of 15 students are: 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30. Find the median. If the weight 44kg is replaced by 46kg and 27kg by 25kg, find the new median.
Answer
Given the numbers are 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30 Arranging the numbers in ascending order 27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45. n = 15 (odd)$\therefore\text{New Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$
$=\Big(\frac{\text{15}+1}{2}\Big)^{\text{th}}\text{value}$
$=8^{\text{th}}\text{value}=35\text{kg}$
If the weight 44kg is replaced by 46kg and 27kg is replaced by 25kg Then the new values be in order are: 25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46$\therefore\text{New Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$
$=\Big(\frac{\text{15}+1}{2}\Big)^{\text{th}}\text{value}$
$=8^{\text{th}}\text{value}=35\text{kg}$
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Question 194 Marks
If the mean of x + 2, 2x + 3, 3x + 4, 4x + 5 is x + 2, find x.
Answer
The given data is x + 2, 2x + 3, 3x + 4, 4x + 5. They are four in numbers. The mean is:$\frac{(\text{x}+2)+(\text{2x}+3)+(\text{3x}+4)+(\text{4x}+5)}{4}$
$=\frac{\text{x}+2+\text{2x}+3+\text{3x}+4+\text{4x}+5}{4}$
$=\frac{\text{10x}+14}{4}$
$=\frac{2(\text{5x}+7)}{4}$
$=\frac{\text{5x}+7}{2}$
But, it is given that the mean is x + 2. Hence, we have$\frac{\text{5x}+7}{2}=\text{x}+2$
$\Rightarrow\text{5x}+7=2(\text{x}+2)$
$\Rightarrow\text{5x}+7=\text{2x}+4$
$\Rightarrow\text{5x}-\text{2x}=4-7$
$\Rightarrow\text{3x}=-3$
$\Rightarrow\text{x}=-\frac{3}{3}$
$\Rightarrow\text{x}=-1$
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Question 204 Marks
If the ratio of mean and median of a certain data is 2 : 3, then find the ratio of its mode and mean.
Answer
Given that the ratio of mean and median of a certain data is 2 : 3. That is, MEAN : MEDIAN = 2 : 3$\Rightarrow\frac{\text{MEAN}}{\text{MEDIAN}}=\frac{2}{3}$
$\Rightarrow3\times\text{MEAN}=2\times\text{MEDIAN}$
$\Rightarrow\text{MEDIAN}=\frac{3}{2}\text{MEAN}$
We know that,$\text{MODE}=3\times\text{MEDIAN}-2\times\text{MEAN}$
$\Rightarrow\text{MODE}=3\times\frac{3}{2}\text{MEAN}-2\times\text{MEAN}$
$\Rightarrow\text{MODE}=\Big(\frac{9}{2}-2\Big)\text{MEAN}$
$\Rightarrow\frac{\text{MODE}}{\text{MEAN}}=\frac{9}{5}-2$
$\Rightarrow\frac{\text{MODE}}{\text{MEAN}}=\frac{5}{2}$
$\Rightarrow\text{MODE}:\text{MEAN}=5:2$
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Question 214 Marks
If the median of $33,28,20,25,34$, $x$ is 29 , find the maximum possible value of $x$.
Answer
The given data is $33,28,20,25,34$, $x$.
The total number of values is $n=6$, is an even number.
Hence the median depends on the $\left(\frac{6}{3}\right)=3^{\text {rd }}$ observation and $\left(\frac{6}{2}+1\right)=4^{\text {th }}$ observation.
Since we have to find the maximum possible value of $x$.
So we must put it in the $4^{\text {th }}$ position when ordering in ascending order.
Arranging the data in ascending order,
we have $20,25,28, x, 33,34$ Hence, the median is:
$\frac{\left(\frac{ n }{2}\right)^{\text {th }} \text { observation }+\left(\frac{ n }{2}+1\right)^{\text {th }} \text { observation }}{2}$
$=\frac{\left(\frac{6}{2}\right)^{\text {th }} \text { observation }+\left(\frac{6}{2}+1\right)^{\text {th }} \text { observation }}{2}$
$=\frac{3^{\text {rd }} \text { observation }+4^{\text {th }} \text { observation }}{2}$
$=\frac{28+x}{2}$
Here it is given that the median is 29 . So, we have $\frac{28+ x }{2}=29$
$\Rightarrow 28+x=58$
$\Rightarrow x=58-28$
$\Rightarrow x=30$
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Question 224 Marks
If the difference of mode and median of a data is 24, then find the difference of median and mean.
Answer
Given that the difference of mode and median of a data is 24. That is, MODE - MEDIAN = 24 MODE = MEDIAN + 24 We have to find the difference between median and mean We know that MODE = 3 × MEDIAN - 2 × MEAN ⇒ MEDIAN + 24 = 3 × MEDIAN - 2 × MEAN ⇒ 24 = 3 × MEDIAN - MEDIAN - 2 × MEAN ⇒ 24 = 2 × MEDIAN - 2 × MEAN ⇒ 2 × MEDIAN - 2 × MEAN = 24 ⇒ 2(MEDIAN - MEAN) = 24$\Rightarrow\text{MEDIAN}-\text{MEAN}=\frac{24}{2}$
⇒ MEDIAN - MEAN = 12
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