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JEE Main 20202 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
2 September
Shift
Morning Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2020 · 2 September · Morning Shift · Mathematics · Statistics

Let X = \ x ∈ N: 1 ≤ x ≤ 17\ and Y = \ ax + b: x ∈ Xanda, b ∈ R, a...

Let X={x∈N:1≤x≤17}X = \left\{x \in N: 1 \le x \le 17\right\} and Y={ax+b:x∈Xanda,b∈R,a>0}Y = \left\{ax + b: x \in Xanda, b \in R, a > 0\right\}. If mean and variance of elements of YY are 1717 and 216216 respectively then a+ba + b is equal to

Options

  1. A77
  2. B−7- 7
  3. C−27- 27
  4. D99

Correct answer

B. −7- 7

Step-by-step solution

Y(x¯)=ax¯+b=a(1+2+3+….+17)17+b=17Y\left(\overset{¯}{x}\right) = a\overset{¯}{x} + b = \dfrac{a\left(1 + 2 + 3 + \ldots . + 17\right)}{17} + b = 17

a(17×18)17×2+b=17\dfrac{a\left(17 \times 18\right)}{17 \times 2} + b = 17

9a+b=179a + b = 17

σ(X2)=∑x2n−(∑xn)2\sigma\left({X}^{2}\right) = \dfrac{∑ {x}^{2}}{n} - {\left(\dfrac{∑ x}{n}\right)}^{2}

=12+22+…+17217−(1+2+...+1717)2= \dfrac{{1}^{2} + {2}^{2} + \ldots + {17}^{2}}{17} - {\left(\dfrac{1 + 2 + . . . + 17}{17}\right)}^{2}

=17⋅18⋅356⋅17−(17⋅182⋅17)2= \dfrac{17 \cdot 18 \cdot 35}{6 \cdot 17} - {\left(\dfrac{17 \cdot 18}{2 \cdot 17}\right)}^{2}

=105−81=24= 105 - 81 = 24

∴σ(Y2)=a2σ(X2)=a2⋅24=216∴ \sigma\left({Y}^{2}\right) = {a}^{2}\sigma\left({X}^{2}\right) = {a}^{2} \cdot 24 = 216

⇒a2=21624=9⇒ {a}^{2} = \dfrac{216}{24} = 9

a=3a = 3

∴b=17−27∴ b = 17 - 27

∴b=17−27∴ b = 17 - 27

b=−10b = - 10

∴a+b=−7∴ a + b = - 7

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