JEE Mains · Maths · STD 11 - 13. statistics
The mean and standard deviation of the marks of \(10\) students were found to be \(50\) and \(12\) respectively. Later, it was observed that two marks \(20\) and \(25\) were wrongly read as \(45\) and \(50\) respectively. Then the correct variance is \(............\).
- A \(265\)
- B \(269\)
- C \(264\)
- D \(289\)
Answer & Solution
Correct Answer
(B) \(269\)
Step-by-step Solution
Detailed explanation
Sol. \(\bar{x}=50\) \(\sum x_i=500\) \(\sum x_{i \text { correct }}=500+20+25-45-50=450\) \(\sigma^2=144\) \(\frac{\sum x_i^2}{10}-(50)^2=144\) \(\sum x_{i c o r r e c t}^2=\left(144+(50)^2\right) \times 10-(45)^2-(50)^2+(20)^2+(25)^2\) \(22940\) Correct variance…
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