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Statistics question

2023 · 8 Apr · Shift 2 · Q27
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  5. /2023 · 8 Apr · Shift 2 · Q27

Statistics question

2023 · 8 Apr · Shift 2 · Q27

JEE MainMathematicsStatisticsMCQ+4 / −1
Let the mean and variance of 12 observations be 92\frac{9}{2}29​ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is mn\frac{m}{n}nm​, where m\mathrm{m}m and n\mathrm{n}n are coprime, then m+n\mathrm{m}+\mathrm{n}m+n is equal to :
  1. A
    317
  2. B
    316
  3. C
    314
  4. D
    315
View written solutionFree

Correct answer: A

  1. Given data for 12 observations
  • Number of observations: n=12n=12n=12
  • गलत (incorrectly calculated) mean: xˉ=92\bar{x}=\dfrac{9}{2}xˉ=29​
  • गलत variance: 444

We use:

Variance=∑xi2n−(∑xin)2\text{Variance} = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2Variance=n∑xi2​​−(n∑xi​​)2
  1. Find the incorrect sum of observations

Since

xˉ=∑xi12=92\bar{x} = \frac{\sum x_i}{12} = \frac{9}{2}xˉ=12∑xi​​=29​

we get

∑xi=12⋅92=54\sum x_i = 12 \cdot \frac{9}{2} = 54∑xi​=12⋅29​=54
  1. Find the incorrect sum of squares

Given variance =4=4=4:

4=∑xi212−(92)24 = \frac{\sum x_i^2}{12} - \left(\frac{9}{2}\right)^24=12∑xi2​​−(29​)2 4=∑xi212−8144 = \frac{\sum x_i^2}{12} - \frac{81}{4}4=12∑xi2​​−481​

So,

∑xi212=4+814=16+814=974\frac{\sum x_i^2}{12} = 4 + \frac{81}{4} = \frac{16+81}{4} = \frac{97}{4}12∑xi2​​=4+481​=416+81​=497​

Hence,

∑xi2=12⋅974=3⋅97=291\sum x_i^2 = 12 \cdot \frac{97}{4} = 3 \cdot 97 = 291∑xi2​=12⋅497​=3⋅97=291
  1. Correct the sum

Two observations were taken as 999 and 101010 instead of correct values 777 and 141414.

So the incorrect sum included 9+10=199+10=199+10=19, while the correct sum should include 7+14=217+14=217+14=21.

Therefore,

Correct sum=54−19+21=56\text{Correct sum} = 54 - 19 + 21 = 56Correct sum=54−19+21=56

Thus correct mean is

xˉcorrect=5612=143\bar{x}_{\text{correct}} = \frac{56}{12} = \frac{14}{3}xˉcorrect​=1256​=314​
  1. Correct the sum of squares

Incorrect contribution of these two observations to sum of squares:

92+102=81+100=1819^2 + 10^2 = 81 + 100 = 18192+102=81+100=181

Correct contribution:

72+142=49+196=2457^2 + 14^2 = 49 + 196 = 24572+142=49+196=245

Therefore,

Correct ∑xi2=291−181+245=355\text{Correct } \sum x_i^2 = 291 - 181 + 245 = 355Correct ∑xi2​=291−181+245=355
  1. Compute the correct variance
σ2=35512−(143)2\sigma^2 = \frac{355}{12} - \left(\frac{14}{3}\right)^2σ2=12355​−(314​)2 =35512−1969= \frac{355}{12} - \frac{196}{9}=12355​−9196​

Taking LCM 363636,

σ2=106536−78436=28136\sigma^2 = \frac{1065}{36} - \frac{784}{36} = \frac{281}{36}σ2=361065​−36784​=36281​

So,

mn=28136\frac{m}{n} = \frac{281}{36}nm​=36281​

where 281281281 and 363636 are coprime.

Hence,

m+n=281+36=317m+n = 281+36 = 317m+n=281+36=317
  1. Option check
  • A: 317317317 ✅
  • B: 316316316 ❌
  • C: 314314314 ❌
  • D: 315315315 ❌

Therefore, the correct option is A.

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