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JEE Mathematics 2026 Question with Solution

The mean and variance of 1010 observations are 99 and 34.234.2, respectively. If 88 of these observations are 2,3,5,10,11,13,15,212, 3, 5, 10, 11, 13, 15, 21, then the mean deviation about the median of all the 1010 observations is:

  • A

    44

  • B

    66

  • C

    55

  • D

    77

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The mean of 1010 observations is 99 and the variance is 34.234.2. Eight observations are 2,3,5,10,11,13,15,212, 3, 5, 10, 11, 13, 15, 21.

Find: The mean deviation about the median of all the 1010 observations.

Use the mean to find the sum of all observations:

Total sum=10×9=90\text{Total sum} = 10 \times 9 = 90

The sum of the given eight observations is

2+3+5+10+11+13+15+21=802 + 3 + 5 + 10 + 11 + 13 + 15 + 21 = 80

Let the remaining two observations be xx and yy. Then

x+y=90−80=10x + y = 90 - 80 = 10

Now use the variance.

σ2=34.2\sigma^2 = 34.2

So the sum of squared deviations from the mean is

∑(xi−μ)2=10×34.2=342\sum (x_i - \mu)^2 = 10 \times 34.2 = 342

For the given eight observations,

(2−9)2+(3−9)2+(5−9)2+(10−9)2+(11−9)2+(13−9)2+(15−9)2+(21−9)2=302(2-9)^2 + (3-9)^2 + (5-9)^2 + (10-9)^2 + (11-9)^2 + (13-9)^2 + (15-9)^2 + (21-9)^2 = 302

Hence,

(x−9)2+(y−9)2=342−302=40(x-9)^2 + (y-9)^2 = 342 - 302 = 40

From x+y=10x + y = 10, write y=10−xy = 10 - x. Substituting,

(x−9)2+(1−x)2=40(x-9)^2 + (1-x)^2 = 40 2x2−20x+42=02x^2 - 20x + 42 = 0 x2−10x+21=0x^2 - 10x + 21 = 0 (x−3)(x−7)=0(x-3)(x-7) = 0

So the two missing observations are 33 and 77.

Therefore, the complete ordered data set is

2,3,3,5,7,10,11,13,15,212, 3, 3, 5, 7, 10, 11, 13, 15, 21

The median is the average of the fifth and sixth terms:

Median=7+102=8.5\text{Median} = \frac{7 + 10}{2} = 8.5

Now compute the mean deviation about the median:

∑∣xi−8.5∣=6.5+5.5+5.5+3.5+1.5+1.5+2.5+4.5+6.5+12.5=40\sum |x_i - 8.5| = 6.5 + 5.5 + 5.5 + 3.5 + 1.5 + 1.5 + 2.5 + 4.5 + 6.5 + 12.5 = 40

Hence,

Mean deviation about median=4010=4\text{Mean deviation about median} = \frac{40}{10} = 4

Therefore, the correct option is A.

Stepwise Extraction of Missing Observations

Given: Mean =9= 9, variance =34.2= 34.2, number of observations =10= 10.

Find: The mean deviation about the median.

First find the missing observations from the sum condition:

x+y=10x + y = 10

Then use the variance condition:

(x−9)2+(y−9)2=40(x-9)^2 + (y-9)^2 = 40

Substitute y=10−xy = 10 - x:

(x−9)2+(1−x)2=40(x-9)^2 + (1-x)^2 = 40

Expanding,

x2−18x+81+x2−2x+1=40x^2 - 18x + 81 + x^2 - 2x + 1 = 40 2x2−20x+42=02x^2 - 20x + 42 = 0 x2−10x+21=0x^2 - 10x + 21 = 0 (x−3)(x−7)=0(x-3)(x-7) = 0

Thus the missing values are 33 and 77.

After arranging all observations,

2,3,3,5,7,10,11,13,15,212, 3, 3, 5, 7, 10, 11, 13, 15, 21

The median is

7+102=8.5\frac{7+10}{2} = 8.5

Now average the absolute deviations from 8.58.5 to get the mean deviation:

4010=4\frac{40}{10} = 4

Therefore, the answer is 44, so the correct option is A.

Common mistakes

  • Using the variance formula incorrectly by taking variance as the sum of squared deviations instead of the mean of squared deviations. Here, you must use ∑(xi−μ)2=nσ2\sum (x_i-\mu)^2 = n\sigma^2, so multiply 34.234.2 by 1010.

  • Finding the median incorrectly by taking the fifth term or sixth term alone. Since there are 1010 observations, the median is the average of the fifth and sixth terms after arranging the data.

  • Computing mean deviation about the mean instead of about the median. The question explicitly asks for deviations from the median, so all absolute deviations must be taken from 8.58.5.

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