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JEE Main 20163 April
Exam
JEE Main
Year
2016
Date
3 April
Subject
Mathematics
Chapter
Statistics

JEE Main 2016 · 3 April · Mathematics · Statistics

If the standard deviation of the numbers 2, 3, a and 11 is 3 . 5 ,...

If the standard deviation of the numbers 2,3,a2, 3, a and 1111 is 3.53 . 5, then which of the following is true ?

Options

  1. A3a2−34a+91=0.3{a}^{2} - 34a + 91 = 0.
  2. B3a2−23a+44=0.3{a}^{2} - 23a + 44 = 0.
  3. C3a2−26a+55=0.3{a}^{2} - 26a + 55 = 0.
  4. D3a2−32a+84=0.3{a}^{2} - 32a + 84 = 0.

Correct answer

D. 3a2−32a+84=0.3{a}^{2} - 32a + 84 = 0.

Step-by-step solution

xx x2{x}^{2}
22 44
33 99
aa a2{a}^{2}
1111 121121
Sum→16+a\text{Sum} \to 16 + a Sum→134+a2\text{Sum} \to 134 + {a}^{2}


σ=∑x24−(∑xi4)2\sigma = \sqrt{\dfrac{∑ {x}^{2}}{4} - {\left(\dfrac{∑ {x}_{i}}{4}\right)}^{2}}

134+a24−(16+a4)2=3510\sqrt{\dfrac{134 + {a}^{2}}{4} - {\left(\dfrac{16 + a}{4}\right)}^{2}} = \dfrac{35}{10}

12134+a2−(16+a)24=72\dfrac{1}{2}\sqrt{134 + {a}^{2} - \dfrac{{\left(16 + a\right)}^{2}}{4}} = \dfrac{7}{2}

536+4a2−256−a2−32a=14\sqrt{536 + 4{a}^{2} - 256 - {a}^{2} - 32a} =14

⇒280+3a2−32a=196⇒ 280 + 3{a}^{2} - 32a = 196

⇒3a2−32a+84=0.⇒ 3{a}^{2} - 32a + 84 = 0.
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