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JEE Main 202325 JanuaryMorning Shift
Exam
JEE Main
Year
2023
Date
25 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2023 · 25 January · Morning Shift · Mathematics · Statistics

The mean and variance of the marks obtained by the students in a...

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:

Options

  1. A4.044 . 04
  2. B4.084 . 08
  3. C3.963 . 96
  4. D3.923 . 92

Correct answer

C. 3.963 . 96

Step-by-step solution

Let the number of observations of the data set is nn.

So,

∑xi=10n....(i)\mathop{∑}_{}{x}_{i} = 10n . . . . \left(i\right)

Also,

∑xi−8+12=10.2n\mathop{∑}_{}{x}_{i} - 8 + 12 = 10 . 2n

⇒10n+4=10.2n⇒ 10n + 4 = 10 . 2n

⇒0.2n=4⇒ 0 . 2n = 4

⇒n=20⇒ n = 20

Also,

σ2=4=∑xi220−100{\sigma}^{2} = 4 = \dfrac{\mathop{∑}_{}{x}_{i}^{2}}{20} - 100

⇒∑xi2=104×20=2080⇒ \mathop{∑}_{}{x}_{i}^{2} = 104 \times 20 = 2080

Now, new variance is

=∑xi2−82+12220−(10.2)2= \dfrac{\mathop{∑}_{}{x}_{i}^{2} - {8}^{2} + {12}^{2}}{20} - {\left(10 . 2\right)}^{2}

=2080+4×2020−104.04= \dfrac{2080 + 4 \times 20}{20} - 104 . 04

=108−104.04= 108 - 104 . 04

=3.96= 3 . 96

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