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JEE Mathematics 2023 Question with Solution

The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and σ2\sigma^2 respectively. If the variance of all the 30 numbers in the two sets is 13, then σ2\sigma^2 is equal to:

  • A

    1212

  • B

    1010

  • C

    1111

  • D

    99

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The first set has 15 numbers with mean 12 and variance 14. The second set has 15 numbers with mean 14 and variance σ2\sigma^2. The variance of all 30 numbers together is 13.

Find: The value of σ2\sigma^2.

Using the relation

σ2=x2‾−(xˉ)2\sigma^2 = \overline{x^2} - (\bar{x})^2

For the first set,

x2‾=14+(12)2=14+144\overline{x^2} = 14 + (12)^2 = 14 + 144

For the second set, let σ2=a\sigma^2 = a. Then

x2‾=a+(14)2=a+196\overline{x^2} = a + (14)^2 = a + 196

For all 30 numbers together, the mean is

xˉ=15⋅12+15⋅1430=13\bar{x} = \frac{15 \cdot 12 + 15 \cdot 14}{30} = 13

Now use the combined variance:

13=(14+144)×15+(a+196)×1530−(13)213 = \frac{(14+144)\times 15 + (a+196)\times 15}{30} - (13)^2

So,

13=15(158+a+196)30−16913 = \frac{15(158 + a + 196)}{30} - 169 13=354+a2−16913 = \frac{354 + a}{2} - 169

Hence,

182=354+a2182 = \frac{354 + a}{2} 364=354+a364 = 354 + a a=10a = 10

Therefore, σ2=10\sigma^2 = 10 and the correct option is B.

Using mean of squares explicitly

The solution uses σ2=∑xi2−(xˉ)2\sigma^2 = \sum x_i^2 - (\bar{x})^2 in mean form, that is,

Variance=x2‾−(xˉ)2\text{Variance} = \overline{x^2} - (\bar{x})^2

Thus the mean of squares for the first set is

14+122=14+144=15814 + 12^2 = 14 + 144 = 158

and for the second set it is

a+142=a+196a + 14^2 = a + 196

Since both sets have equal size 15, the combined mean of squares is the average of these two values:

x2‾combined=158+(a+196)2\overline{x^2}_{\text{combined}} = \frac{158 + (a+196)}{2}

Also, the combined mean is

12+142=13\frac{12+14}{2} = 13

Therefore,

13=x2‾combined−13213 = \overline{x^2}_{\text{combined}} - 13^2 13=158+a+1962−16913 = \frac{158 + a + 196}{2} - 169

Solving gives

a=10a = 10

So, σ2=10\sigma^2 = 10.

Common mistakes

  • Using the combined variance as the average of the two variances, 14+σ22\frac{14 + \sigma^2}{2}, is wrong because the means of the two sets are different. First account for the shift in means through x2‾−(xˉ)2\overline{x^2} - (\bar{x})^2.

  • Forgetting to compute the combined mean before using the variance formula is incorrect. The variance of all 30 numbers must use the overall mean, which here is 1313.

  • Using 14214^2 for the first set or 12212^2 for the second set by interchanging the means is a setup error. Use mean 12 with variance 14 for the first set and mean 14 with variance σ2\sigma^2 for the second set.

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