JEE Mains · Maths · STD 11 - 13. statistics
Let \(\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{N}\) and \(\mathrm{a}<\mathrm{b}<\mathrm{c}\). Let the mean, the mean deviation about the mean and the variance of the \(5\) observations \(9\),\(25\), \(a\), \(b\), \(c\) be \(18\),\(4\) and \(\frac{136}{5}\), respectively. Then \(2 \mathrm{a}+\mathrm{b}-\mathrm{c}\) is equal to ..............
- A \(39\)
- B \(18\)
- C \(35\)
- D \(33\)
Answer & Solution
Correct Answer
(D) \(33\)
Step-by-step Solution
Detailed explanation
\( a, b, c \in N \quad a < b < c \) \( \bar{x}=\text { mean }=\frac{9+25+a+b+c}{5}=18 \) \( a+b+c=56 \) \( \text { Mean deviation }=\frac{\sum\left|x_i-\bar{x}\right|}{n}=4 \) \( =9+7+|18-a|+|18-b|+|18-c|=20 \) \( =|18-a|+|18-b|+|18-c|=4 \)…
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