यदि x प्रायिकता घनत्व फलन के साथ एक सतत यादृच्छिक चर है

f(x) = 1/√2πe-x2/2dx <x<

और y को y = x + 1 के रूप में परिभाषित किया गया है, तो E(y) किसके बराबर है?

This question was previously asked in
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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Detailed Solution

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दिया गया है

f(x) = 1/√2πe-x2/2dx

सूत्र

E(y)=yf(x)dx

गणना 

E(y)=yf(x)dx=(x+1)1/2π(e-x2/2dx

⇒ (1/√2π)(x+1)e-x2/2dx

⇒ E(y) = (1/√2π)x × e-x2/2dx + (1/√2π)(e-x2/2dx

⇒ चूंकि, x × e-x2/2dx k का विषम फलन है

∴ (1/2√2π)(e-x2/2dx = 0

⇒ E(y) = (1/2√2π)(e-x2/2dx

⇒ (1/√2π) × √2π = 1

मानक समाकलन का प्रयोग करने पर हमारे पास है 

e-x2/2dx = √2π 

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