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JEE Main 202427 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
27 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2024 · 27 January · Morning Shift · Mathematics · Statistics

Let a _ 1 , a _ 2 , . . . , a _ 10 be 10 observations such that ∑ _...

Let a1,a2,...,a10{a}_{1}, {a}_{2}, . . . , {a}_{10} be 1010 observations such that ∑k=110ak=50\mathop{∑}_{k = 1}^{10}{a}_{k} = 50 and ∑∀k<jak⋅aj=1100\mathop{∑}_{\forall k < j}{a}_{k} \cdot {a}_{j} = 1100. Then the standard deviation of a1,a2,…,a10{a}_{1}, {a}_{2}, \ldots , {a}_{10} is equal to :

Options

  1. A55
  2. B5\sqrt{5}
  3. C1010
  4. D115\sqrt{115}

Correct answer

B. 5\sqrt{5}

Step-by-step solution

Given: ∑k=110ak=50\mathop{∑}_{k = 1}^{10}{a}_{k} = 50

⇒a1+a2+…+a10=50...(i)⇒ {a}_{1} + {a}_{2} + \ldots + {a}_{10} = 50 . . . \left(i\right)

⇒(a1+a2+…+a10)2=2500⇒ {\left({a}_{1} + {a}_{2} + \ldots + {a}_{10}\right)}^{2} = 2500

⇒∑i=110ai2+2∑k<jakaj=2500⇒ \mathop{∑}_{i = 1}^{10}{a}_{i}^{2} + 2\mathop{∑}_{k < j}{a}_{k}{a}_{j} = 2500

⇒∑i=110ai2+2×1100=2500⇒ \mathop{∑}_{i = 1}^{10}{a}_{i}^{2} + 2 \times 1100 = 2500

⇒∑i=110ai2=2500−2(1100)⇒ \mathop{∑}_{i = 1}^{10}{a}_{i}^{2} = 2500 - 2\left(1100\right)

⇒∑i=110ai2=300⇒ \mathop{∑}_{i = 1}^{10}{a}_{i}^{2} = 300

 Standard deviation, σ=∑ai2n−(∑ain)2\sigma = \sqrt{\dfrac{∑ {a}_{i}^{2}}{n} - {\left(\dfrac{∑ {a}_{i}}{n}\right)}^{2}}

⇒σ=∑ai210−(∑ai10)2⇒ \sigma = \sqrt{\dfrac{∑ {a}_{i}^{2}}{10} - {\left(\dfrac{∑ {a}_{i}}{10}\right)}^{2}}

⇒σ=30010−(5010)2⇒ \sigma = \sqrt{\dfrac{300}{10} - {\left(\dfrac{50}{10}\right)}^{2}}

⇒σ=30−25=5⇒ \sigma = \sqrt{30 - 25} = \sqrt{5}

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