JEE Mains · Maths · STD 11 - 13. statistics
Let \(X=\{\mathrm{x} \in \mathrm{N}: 1 \leq \mathrm{x} \leq 17\}\) and \(\mathrm{Y}=\{\mathrm{ax}+\mathrm{b}: \mathrm{x} \in \mathrm{X}\) and \(\mathrm{a}, \mathrm{b} \in \mathrm{R}, \mathrm{a}>0\} .\) If mean and variance of elements of \(Y\) are \(17\) and \(216\) respectively then \(a + b\) is equal to
- A \(-7\)
- B \(7\)
- C \(9\)
- D \(-27\)
Answer & Solution
Correct Answer
(A) \(-7\)
Step-by-step Solution
Detailed explanation
\(\sigma^{2}=\) variance \(\mu=\) mean \(\sigma^{2}=\frac{\sum_{i=1}^{n}\left(x_{i}-\mu\right)^{2}}{n}\) \(\mu=17\) \(\Rightarrow \frac{\sum_{x=1}^{17}(a x+b)}{17}=17\) \(\Rightarrow \quad 9 a+b=17\) \(\sigma^{2}=216\)…
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