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Statistics question

2016 · 10 Apr · Shift 1 · Q31
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Statistics question

2016 · 10 Apr · Shift 1 · Q31

JEE MainMathematicsStatisticsMCQ+4 / −1
The mean of 5 observations is 5 and their variance is 124. If three of the observations are 1, 2 and 6 ; then the mean deviation from the mean of the data is :
  1. A
    2.4
  2. B
    2.8
  3. C
    2.5
  4. D
    2.6
View written solutionFree

Correct answer: QUESTION/DATA APPEARS INCONSISTENT. USING THE GIVEN MEAN $5$ AND VARIANCE $124$, THE MEAN DEVIATION FROM MEAN IS $\FRAC{8+24\SQRT2}{5}\APPROX 8.39$, SO OPTION B IS NOT CORRECT.

  1. Use the mean to find the sum of all 5 observations

Given mean =5=5=5 for 555 observations,

sum of observations=5×5=25.\text{sum of observations} = 5\times 5 = 25.sum of observations=5×5=25.

Three observations are 1,2,61,2,61,2,6. Let the remaining two be xxx and yyy. Then

1+2+6+x+y=251+2+6+x+y=251+2+6+x+y=25 9+x+y=259+x+y=259+x+y=25 x+y=16.x+y=16.x+y=16.
  1. Use the variance to find xxx and yyy

For JEE-type questions, variance of the data is

σ2=1n∑(xi−xˉ)2.\sigma^2=\frac{1}{n}\sum (x_i-\bar x)^2.σ2=n1​∑(xi​−xˉ)2.

Given variance =124=124=124 and mean xˉ=5\bar x=5xˉ=5,

15∑(xi−5)2=124\frac{1}{5}\sum (x_i-5)^2=12451​∑(xi​−5)2=124 ∑(xi−5)2=620.\sum (x_i-5)^2=620.∑(xi​−5)2=620.

Now for the known observations:

(1−5)2=16,(1-5)^2=16,(1−5)2=16, (2−5)2=9,(2-5)^2=9,(2−5)2=9, (6−5)2=1.(6-5)^2=1.(6−5)2=1.

Their sum is

16+9+1=26.16+9+1=26.16+9+1=26.

So,

(x−5)2+(y−5)2=620−26=594.(x-5)^2+(y-5)^2=620-26=594.(x−5)2+(y−5)2=620−26=594.

Expand:

(x−5)2+(y−5)2=x2+y2−10(x+y)+50.(x-5)^2+(y-5)^2=x^2+y^2-10(x+y)+50.(x−5)2+(y−5)2=x2+y2−10(x+y)+50.

Since x+y=16x+y=16x+y=16,

x2+y2−160+50=594x^2+y^2-160+50=594x2+y2−160+50=594 x2+y2=704.x^2+y^2=704.x2+y2=704.

Now,

(x+y)2=x2+y2+2xy(x+y)^2=x^2+y^2+2xy(x+y)2=x2+y2+2xy 162=704+2xy16^2=704+2xy162=704+2xy 256=704+2xy256=704+2xy256=704+2xy 2xy=−448Rightarrowxy=−224.2xy=-448 Rightarrow xy=-224.2xy=−448Rightarrowxy=−224.

Thus x,yx,yx,y satisfy

t2−16t−224=0.t^2-16t-224=0.t2−16t−224=0.

Solving,

t=8±418=8±122.t=8\pm 4\sqrt{18}=8\pm 12\sqrt{2}.t=8±418​=8±122​.

So the two observations are

8+122,8−122.8+12\sqrt{2},\quad 8-12\sqrt{2}.8+122​,8−122​.
  1. Compute mean deviation from the mean

Mean deviation from mean is

15∑∣xi−xˉ∣.\frac{1}{5}\sum |x_i-\bar x|.51​∑∣xi​−xˉ∣.

Here xˉ=5\bar x=5xˉ=5, so

MD=15(∣1−5∣+∣2−5∣+∣6−5∣+∣8+122−5∣+∣8−122−5∣).\text{MD} = \frac{1}{5}\left(|1-5|+|2-5|+|6-5|+|8+12\sqrt2-5|+|8-12\sqrt2-5|\right).MD=51​(∣1−5∣+∣2−5∣+∣6−5∣+∣8+122​−5∣+∣8−122​−5∣).

Now,

∣1−5∣=4,∣2−5∣=3,∣6−5∣=1.|1-5|=4, \quad |2-5|=3, \quad |6-5|=1.∣1−5∣=4,∣2−5∣=3,∣6−5∣=1.

Also,

∣8+122−5∣=∣3+122∣=3+122,|8+12\sqrt2-5|=|3+12\sqrt2|=3+12\sqrt2,∣8+122​−5∣=∣3+122​∣=3+122​, ∣8−122−5∣=∣3−122∣=122−3|8-12\sqrt2-5|=|3-12\sqrt2|=12\sqrt2-3∣8−122​−5∣=∣3−122​∣=122​−3

since 122>312\sqrt2>3122​>3.

Therefore,

MD=15(4+3+1+(3+122)+(122−3))\text{MD} = \frac{1}{5}\left(4+3+1+(3+12\sqrt2)+(12\sqrt2-3)\right)MD=51​(4+3+1+(3+122​)+(122​−3)) =15(8+242).=\frac{1}{5}(8+24\sqrt2).=51​(8+242​). MD=8+2425.\text{MD}=\frac{8+24\sqrt2}{5}.MD=58+242​​.

Numerically,

MD≈8+24(1.414)5=8+33.9365=41.9365≈8.39.\text{MD}\approx \frac{8+24(1.414)}{5} =\frac{8+33.936}{5} =\frac{41.936}{5} \approx 8.39.MD≈58+24(1.414)​=58+33.936​=541.936​≈8.39.
  1. Compare with the options

The computed mean deviation is approximately

8.39,8.39,8.39,

which does not match any of the given options 2.4,2.8,2.5,2.62.4, 2.8, 2.5, 2.62.4,2.8,2.5,2.6.

So the data in the question is inconsistent with the options. Most likely, the stated variance 124124124 is incorrect.

  1. Check against stored correct answer

Stored correct answer is B: 2.8, but this does not agree with the value obtained from the given data.

Hence, I disagree with the stored answer.

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