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JEE Mathematics 2024 Question with Solution

If the mean and variance of five observations are 245\frac{24}{5} and 19425\frac{194}{25} respectively, and the mean of the first four observations is 72\frac{7}{2}, then the variance of the first four observations is equal to:

  • A

    45\frac{4}{5}

  • B

    7712\frac{77}{12}

  • C

    54\frac{5}{4}

  • D

    1054\frac{105}{4}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The mean of five observations is 245\frac{24}{5} and their variance is 19425\frac{194}{25}. The mean of the first four observations is 72\frac{7}{2}.

Find: The variance of the first four observations.

Let the five observations be x1,x2,x3,x4,x5x_1, x_2, x_3, x_4, x_5.

From the mean of five observations,

x1+x2+x3+x4+x55=245\frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5}

so,

x1+x2+x3+x4+x5=24x_1+x_2+x_3+x_4+x_5=24

From the mean of the first four observations,

x1+x2+x3+x44=72\frac{x_1+x_2+x_3+x_4}{4}=\frac{7}{2}

so,

x1+x2+x3+x4=14x_1+x_2+x_3+x_4=14

Hence, the fifth observation is

x5=24−14=10x_5=24-14=10

Using the variance formula for five observations,

σ2=∑i=15(xi−xˉ)25=19425\sigma^2=\frac{\sum_{i=1}^{5}(x_i-\bar{x})^2}{5}=\frac{194}{25}

where

xˉ=245\bar{x}=\frac{24}{5}

So,

∑i=15(xi−245)2=5⋅19425=1945\sum_{i=1}^{5}(x_i-\tfrac{24}{5})^2=5\cdot \frac{194}{25}=\frac{194}{5}

Now separate the contribution of x5=10x_5=10:

(x5−245)2=(10−245)2=(265)2=67625(x_5-\tfrac{24}{5})^2=\left(10-\frac{24}{5}\right)^2=\left(\frac{26}{5}\right)^2=\frac{676}{25}

Therefore,

∑i=14(xi−245)2=1945−67625=970−67625=29425\sum_{i=1}^{4}(x_i-\tfrac{24}{5})^2=\frac{194}{5}-\frac{676}{25}=\frac{970-676}{25}=\frac{294}{25}

Let the mean of the first four observations be

yˉ=72\bar{y}=\frac{7}{2}

Using the shift of origin relation,

∑i=14(xi−yˉ)2=∑i=14(xi−245)2−4(yˉ−245)2\sum_{i=1}^{4}(x_i-\bar{y})^2=\sum_{i=1}^{4}(x_i-\tfrac{24}{5})^2-4\left(\bar{y}-\frac{24}{5}\right)^2

Now,

yˉ−245=72−245=35−4810=−1310\bar{y}-\frac{24}{5}=\frac{7}{2}-\frac{24}{5}=\frac{35-48}{10}=-\frac{13}{10}

so,

4(yˉ−245)2=4⋅169100=169254\left(\bar{y}-\frac{24}{5}\right)^2=4\cdot \frac{169}{100}=\frac{169}{25}

Hence,

∑i=14(xi−yˉ)2=29425−16925=12525=5\sum_{i=1}^{4}(x_i-\bar{y})^2=\frac{294}{25}-\frac{169}{25}=\frac{125}{25}=5

Therefore, the variance of the first four observations is

54\frac{5}{4}

So, the correct option is C.

Common mistakes

  • Using the mean 245\frac{24}{5} of all five observations as the mean of the first four observations is incorrect. The first four observations have a different mean, namely 72\frac{7}{2}, which must be used for their variance.

  • Finding x5=10x_5=10 and then stopping is incomplete. The fifth observation helps separate the total variance contribution, but one still has to convert the spread about 245\frac{24}{5} into the spread about 72\frac{7}{2} for the first four observations.

  • Applying the variance formula with denominator 55 for the first four observations is wrong. Since there are only four observations in the required set, the correct denominator is 44.

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