If x is a continuous random variable with p.d.f.
f(x) = 1/√2πe-x2/2dx \( - ∞<x<∞\)

and y is defined as y = x + 1, then E(y) equals:

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SSC CGL Tier-II ( JSO ) 2016 Official Paper ( Held On : 2 Dec 2016 )
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  1. 1
  2. 0
  3. n
  4. (√n) + 1

Answer (Detailed Solution Below)

Option 1 : 1
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PYST 1: SSC CGL - English (Held On : 11 April 2022 Shift 1)
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Detailed Solution

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Given

f(x) = 1/√2πe-x2/2dx

Formula

\(E\left( y \right) = \;\mathop \smallint \limits_{ - \infty }^\infty yf\left( x \right)dx\)

Calculation

\(E\left( y \right) = \;\mathop \smallint \limits_{ - \infty }^\infty yf\left( x \right)dx = \;\mathop \smallint \limits_{ - \infty }^\infty \left( {x + 1} \right)1/√ {2π } \)(e-x2/2dx

⇒ (1/√2π)\(\mathop \smallint \nolimits_{ - \infty }^\infty \left( {x + 1} \right)\)e-x2/2dx

⇒ E(y) = (1/√2π)\(\mathop \smallint \limits_{ - \infty }^\infty \)x × e-x2/2dx + (1/√2π)(\(\mathop \smallint \limits_{ - \infty }^\infty \)e-x2/2dx

⇒ Since, x × e-x2/2dx is an odd function of k 

∴ (1/2√2π)(\(\mathop \smallint \limits_{ - \infty }^\infty \)e-x2/2dx = 0

⇒ E(y) = (1/2√2π)(\(\mathop \smallint \limits_{ - \infty }^\infty \)e-x2/2dx

⇒ (1/√2π) × √2π = 1

Using the standard integrals we have

\(\mathop \smallint \limits_{ - \infty }^\infty \)e-x2/2dx  = √2π 

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