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JEE Mathematics 2023 Question with Solution

Let the mean of 66 observations 1,2,4,5,x,1, 2, 4, 5, x, and yy and their variance be 1010. Then their mean deviation about the mean is equal to:

  • A

    73\frac{7}{3}

  • B

    103\frac{10}{3}

  • C

    83\frac{8}{3}

  • D

    33

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The observations are 1,2,4,5,x,y1, 2, 4, 5, x, y. Their mean is 55 and variance is 1010.

Find: The mean deviation about the mean.

Using the mean condition:

1+2+4+5+x+y6=5\frac{1+2+4+5+x+y}{6}=5

So,

12+x+y6=5\frac{12+x+y}{6}=5 12+x+y=3012+x+y=30 x+y=18x+y=18

Using the variance formula:

12+22+42+52+x2+y26−52=10\frac{1^2+2^2+4^2+5^2+x^2+y^2}{6}-5^2=10 46+x2+y26−25=10\frac{46+x^2+y^2}{6}-25=10 46+x2+y26=35\frac{46+x^2+y^2}{6}=35 46+x2+y2=21046+x^2+y^2=210 x2+y2=164x^2+y^2=164

Now use

(x+y)2=x2+y2+2xy(x+y)^2=x^2+y^2+2xy

Substituting the values:

182=164+2xy18^2=164+2xy 324=164+2xy324=164+2xy 2xy=1602xy=160 xy=80xy=80

Hence xx and yy are roots of

t2−(x+y)t+xy=0t^2-(x+y)t+xy=0 t2−18t+80=0t^2-18t+80=0

Factoring,

(t−8)(t−10)=0(t-8)(t-10)=0

So, x=8x=8 and y=10y=10, or vice versa.

Now compute the mean deviation about the mean 55:

Mean Deviation=∣1−5∣+∣2−5∣+∣4−5∣+∣5−5∣+∣8−5∣+∣10−5∣6\text{Mean Deviation}=\frac{|1-5|+|2-5|+|4-5|+|5-5|+|8-5|+|10-5|}{6} =4+3+1+0+3+56=\frac{4+3+1+0+3+5}{6} =166=83=\frac{16}{6}=\frac{8}{3}

Therefore, the mean deviation about the mean is 83\frac{8}{3}. The correct option is C.

Using sum and sum of squares

Given: Mean =5=5, variance =10=10 for the six observations 1,2,4,5,x,y1,2,4,5,x,y.

Find: Mean deviation about the mean.

From the mean, total sum of observations is

6×5=306\times 5=30

The known four numbers add up to

1+2+4+5=121+2+4+5=12

Therefore,

x+y=30−12=18x+y=30-12=18

From the variance relation,

∑xi26−(5)2=10\frac{\sum x_i^2}{6}-(5)^2=10

So,

∑xi26=35\frac{\sum x_i^2}{6}=35

Hence,

∑xi2=210\sum x_i^2=210

Now the known squares sum to

12+22+42+52=461^2+2^2+4^2+5^2=46

Thus,

x2+y2=210−46=164x^2+y^2=210-46=164

Use

(x+y)2=x2+y2+2xy(x+y)^2=x^2+y^2+2xy

which gives

182=164+2xy18^2=164+2xy 324=164+2xy324=164+2xy xy=80xy=80

Therefore the pair is obtained from

t2−18t+80=0t^2-18t+80=0

so the values are 88 and 1010.

The observations are therefore 1,2,4,5,8,101,2,4,5,8,10 with mean 55. Their absolute deviations from the mean are 4,3,1,0,3,54,3,1,0,3,5. Hence,

Mean Deviation=4+3+1+0+3+56=166=83\text{Mean Deviation}=\frac{4+3+1+0+3+5}{6}=\frac{16}{6}=\frac{8}{3}

So the correct option is C.

Common mistakes

  • Using variance as ∑(xi−xˉ)2n−1\frac{\sum (x_i-\bar{x})^2}{n-1} instead of ∑(xi−xˉ)2n\frac{\sum (x_i-\bar{x})^2}{n} is wrong here because the question uses variance of the full set of observations, not sample variance. Use denominator 66, not 55.

  • Forgetting that the given mean is 55 and treating it as 1010 is incorrect. The statement says the variance is 1010, while the solution shows the mean is 55 from the data.

  • Computing mean deviation without first finding xx and yy is wrong because the full data set is needed. First use the mean and variance conditions to determine x+yx+y and x2+y2x^2+y^2, then find xx and yy.

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