RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download
Q 1 . Calculate the mean for the following distribution :
| x : | 5 | 6 | 7 | 8 | 9 |
| f : | 4 | 8 | 14 | 11 | 3 |
SOLUTION :
| x | f | fx |
| 5 6 7 8 9 | 4 8 14 11 3 | 20 48 98 88 27 |
| N=40 | ∑fx = 281 |
Q 2 . Find the mean of the following data :
| x : | 19 | 21 | 23 | 25 | 27 | 29 | 31 |
| f : | 13 | 15 | 16 | 18 | 16 | 15 | 13 |
SOLUTION :
| x | f | fx |
| 19 21 23 25 27 29 31 | 13 15 16 18 16 15 13 | 247 315 368 450 432 435 403 |
| N=106 | ∑fx=2650 |
Q 3 . 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 |
SOLUTION :
| x | f | fx |
| 10 15 P 25 35 | 3 10 25 7 5 | 30 150 25p 175 175 |
| N = 50 | ∑fx=25p+530 |
It is given that ,
Mean = 20.6
⇒ 25p+530 = 20.6×50
⇒ 25p = 1030−530
⇒ 25p = 500
⇒ p = 50025 = 20
⇒ p = 20
∴ p = 20.
Q 4 . If the mean of the following data is 15 , find p.
| x : | 5 | 10 | 15 | 20 | 25 |
| f : | 6 | p | 6 | 10 | 5 |
SOLUTION :
| x | f | fx |
| 5 10 15 20 25 | 6 P 6 10 5 | 30 10p 90 200 125 |
| N = p+27 | ∑fx = 10p+445 |
It is given that ,
Mean = 15
⇒ 10p+445 = 15×(p+27)
⇒10p+445 = 15p+405
⇒15p−10p = 445−405
⇒ 5p = 40
⇒ p = 40/5 =8
⇒ p = 8
∴ p = 8.
Q 5. 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 |
SOLUTION :
| x | f | fx |
| 8 12 15 P 20 25 30 | 12 16 20 24 16 8 4 | 96 192 300 24p 320 200 120 |
| N = 100 | ∑fx = 24p+1228 |
It is given that ,
Mean = 16.6
⇒ 24p+1228 = 1660
⇒ 24p = 1660−1228
⇒ 24p = 432
⇒ p = 432/24 = 18
⇒ p = 18
∴ p = 18.
Q 6 . 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 |
SOLUTION :
| x | f | fx |
| 5 8 10 12 P 20 25 | 2 5 8 22 7 4 2 | 10 40 80 264 7p 80 50 |
| N = 50 | ∑fx=7p+524 |
It is given that ,
Mean = 12.58
⇒ 7p+524 = 629
⇒ 7p = 629−524
⇒ 7p = 105
⇒ p = 1057 = 15
⇒ p = 15
∴ p = 18.
Q7 . 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 |
SOLUTION :
| x | f | fx |
| 3 5 7 9 11 13 | 6 8 15 P 8 4 | 18 40 105 9p 88 52 |
| N=p+41 | ∑fx=9p+303 |
It is given that ,
Mean = 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
∴ p = 9.
Q 8. 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 |
SOLUTION :
| x | f | fx |
| 15 17 19 20+p 23 | 2 3 4 5p 6 | 30 51 76 100p+ 5p2 138 |
| N=5p+15 | fx= 5p2+100p+295 |
It is given that ,
Mean = 20
⇒ 5p2+100p+295 = 20(5p+15)
⇒ 5p2+100p+295 = 100p+300
⇒ 5p2 =300−295
⇒ 5p2 = 5
⇒ p2 = 1
⇒ p = ±1
Frequency can’t be negative.
Hence, value of p is 1.
Q 9 . Find the mean of the following distribution :
| x : | 10 | 12 | 20 | 25 | 30 |
| f : | 3 | 10 | 15 | 7 | 5 |
SOLUTION :
| x | f | fx |
| 10 12 20 25 35 | 3 10 15 7 5 | 30 120 300 175 175 |
| N = 40 | ∑fx = 800 |
Q 10. Candidates of four schools appear in a mathematics test. The data were as follows :
| Schools | No. Of Candidates | Average Score |
| I II III IV | 60 48 Not Available 40 | 75 80 55 50 |
If the average score of the candidates of all four schools is 66 , Find the number of candidates that appeared from school III .
SOLUTION :
| Schools | No. Of Candidates | Average Score |
| I II III IV | 60 48 x 40 | 75 80 55 50 |
Given the average score of all schools = 66
⇒ 10340+55x = 66x+9768
⇒ 10340−9768 = 66x−55x
⇒ 11x = 572
⇒ x = 572/11 = 52
∴ No. of candidates appeared from school III = 52.
Q.11 . 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 1 2 3 4 5 | 38 144 342 287 164 25 |
| Total | 1000 |
SOLUTION :
| No . of heads per toss(x) | No.of tosses(f) | fx |
| 0 1 2 3 4 5 | 38 144 342 287 164 25 | 0 144 684 861 656 125 |
| N = 1000 | ∑fx = 2470 |
∴ Mean number of heads per toss
= 2470/1000
= 2.47
Q 12 . 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 | f1 | 32 | f2 | 19 |
Total = 120
SOLUTION :
| x | f | fx |
| 10
30 50 70 90 | 17
f1 32 f2 19 | 170
30f1 1600 70f2 1710 |
| N=120 | ∑fx = 3480+30f1+70f2 |
It is given that
Mean = 50
⇒ 3480+30f1+70f2 = 50×120
⇒ 30f1+70f2 = 6000−3480
⇒10(3f1+7f2) = 10(252)
⇒3f1+7f2 = 252⋅⋅⋅⋅⋅(1) [∵Divideby10]
And N = 20
⇒17+f1+32+f2+19 = 120
⇒ 68+f1+f2 = 120
⇒ f1+f2 = 120−68
⇒ f1+f2 = 52
Multiply with 3 on both sides
⇒3f1+3f2 = 156⋅⋅⋅⋅⋅⋅(2)
Subtracting equation (2) from equation (1)
⇒3f1+7f2−3f1−3f2 = 252−156
⇒ 4f2=96
⇒ f2 = 96/4 = 24
Put the value of f2 in equation (1)
⇒ 3f1+7×24 = 252
⇒ 3f1 = 252−168
⇒ f1 = 84/3 = 28
⇒ f1 = 28
FAQs on RD Sharma Solutions Ex-24.2, Measures Of Central Tendency, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics
| 1. What are measures of central tendency in statistics? |  |
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| 3. What is the median and how is it calculated? | |
| 4. How is the mode determined? | |
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Nov 22, 2024 Last updated |
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