1.

The probability density function (PDF) of a random variable, X is given by,fX(x)=Ke−(x+1)218The value of K will be______.0.133

Answer» The probability density function (PDF) of a random variable, X is given by,

fX(x)=Ke(x+1)218

The value of K will be______.
  1. 0.133


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