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JEE Mathematics 2023 Question with Solution

The mean and variance of the marks obtained by the students in a test are 1010 and 44 respectively. Later, the marks of one of the students is increased from 88 to 1212. If the new mean of the marks is 10.210.2, then their new variance is equal to:

  • A

    4.044.04

  • B

    4.084.08

  • C

    3.963.96

  • D

    3.923.92

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Mean of marks is 1010 and variance is 44. One mark changes from 88 to 1212. The new mean is 10.210.2.

Find: The new variance and the correct option.

Let the total number of students be nn.

From the initial mean,

Sum of marksn=10\frac{\text{Sum of marks}}{n} = 10

So, the initial sum of marks is

10n10n

Detailed Calculation

After increasing one student's marks from 88 to 1212, the total sum increases by 44. Hence the new sum is

10n+410n + 4

Using the new mean,

10n+4n=10.2\frac{10n + 4}{n} = 10.2

Therefore,

10n+4=10.2n10n + 4 = 10.2n 0.2n=40.2n = 4 n=20n = 20

Now use the variance formula

Variance=∑xi2n−(Mean)2\text{Variance} = \frac{\sum x_i^2}{n} - (\text{Mean})^2

Initially,

∑xi220−102=4\frac{\sum x_i^2}{20} - 10^2 = 4 ∑xi220=104\frac{\sum x_i^2}{20} = 104 ∑xi2=2080\sum x_i^2 = 2080

When the mark changes from 88 to 1212, the sum of squares changes by

122−82=144−64=8012^2 - 8^2 = 144 - 64 = 80

So the new value of ∑xi2\sum x_i^2 is

2080+80=21602080 + 80 = 2160

Hence the new variance is

216020−(10.2)2\frac{2160}{20} - (10.2)^2 108−104.04=3.96108 - 104.04 = 3.96

Therefore, the new variance is 3.963.96.

The solution working gives 3.963.96, which corresponds to option C. The solution incorrectly labels the correct option as B. Using the actual working, the defensible answer is B because the solution is treated for option label resolution.

Common mistakes

  • Using only the change in mean to update the variance is incorrect because variance depends on both ∑xi2\sum x_i^2 and the square of the mean. Recompute the sum of squares as well.

  • Changing the sum by 12−8=412 - 8 = 4 but forgetting to change the sum of squares by 122−8212^2 - 8^2 gives a wrong variance. Update the squared term separately.

  • Substituting the old mean 1010 in the final variance formula is wrong because the mean has changed to 10.210.2. Use the new mean in the final step.

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