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Mathematics · Probability

JEE Main 2025 — 24 January, Morning Shift — Question 14

For a statistical data x1,x2,…,x10x_{1}, x_{2}, \ldots, x_{10} of 10 values, a student obtained the mean as 5.5 and ∑i=110xi2=371\sum_{i=1}^{10} \mathrm{x}_{\mathrm{i}}^{2}=371. He later found that he had noted two values in the data incorrectly as 4 and 5 , instead of the correct values 6 and 8 , respectively. The variance of the corrected data is

  1. Option A:

    7

    Correct
  2. Option B:

    4

  3. Option C:

    9

  4. Option D:

    5

Answer: A

Step-by-step solution

Mean x‾=5.5\overline{\mathrm{x}}=5.5

=∑i=110xi=5.5×10=55=∑i=110xi2=371(∑xi)new =55−(4+5)+(6+8)=60(∑xi2)new =371−(42+52)+(62+82)=430\begin{aligned} = & \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}=5.5 \times 10=55\\ = & \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}^{2}=371\\ & \left(\sum \mathrm{x}_{\mathrm{i}}\right)_{\text {new }}=55-(4+5)+(6+8)=60\\ & \left(\sum \mathrm{x}_{\mathrm{i}}^{2}\right)_{\text {new }}=371-\left(4^{2}+5^{2}\right)+\left(6^{2}+8^{2}\right)=430 \end{aligned}  Variance (σ2)=∑xi210−(∑xi10)2\text { Variance } (\sigma^{2})=\frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{10}-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{10}\right)^{2} σ2=43010−(6010)2\sigma^{2}=\frac{430}{10}-\left(\frac{60}{10}\right)^{2} σ2=43−36\sigma^{2}=43-36 σ2=7\sigma^{2}=7

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
24 January, Morning Shift
Subject
Mathematics
Chapter
Probability
Topic
Mean, variance, expected values of distributions

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