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I'm given a pdf in a different form and asked to identify the random variable. Also, there is a positive constant $c$ that i have to identify.

PDF: $f(x)= c*\exp(-x^2-x/4)\ \forall -\infty<x<\infty $$f(x)= c*\exp(-x^2-x/4)\ \forall x\in\mathbb{R} $

I suspect this is the normal distribution but I am not sure.

I have tried to factor the exponent by adding and subtracting 1/64

$$ \begin{align} c\ exp(-x^2-x/4)&=c\ exp(-(x^2+x/4+1/64-1/64))\\ &=c\ exp(-(x^2+x/4+1/64)+1/64)\\ &=c\ exp(-(x+1/8)(x+1/8)+1/64)\\ &=c\ exp(-(x+1/8)(x+1/8))*exp(1/64)\\ &=c*exp(1/64)exp\big(-(x+1/8)(x+1/8)\big) \end{align} $$$$ \begin{align} c\ \exp(-x^2-x/4)&=c\ \exp(-(x^2+x/4+1/64-1/64))\\ &=c\ \exp(-(x^2+x/4+1/64)+1/64)\\ &=c\ \exp(-(x+1/8)(x+1/8)+1/64)\\ &=c\ \exp(-(x+1/8)(x+1/8))*\exp(1/64)\\ &=c\exp(1/64)\exp\big(-(x+1/8)(x+1/8)\big) \end{align} $$

but I keep getting stuck. Does anyone have any ideas?

I'm given a pdf in a different form and asked to identify the random variable. Also, there is a positive constant $c$ that i have to identify.

PDF: $f(x)= c*\exp(-x^2-x/4)\ \forall -\infty<x<\infty $

I suspect this is the normal distribution but I am not sure.

I have tried to factor the exponent by adding and subtracting 1/64

$$ \begin{align} c\ exp(-x^2-x/4)&=c\ exp(-(x^2+x/4+1/64-1/64))\\ &=c\ exp(-(x^2+x/4+1/64)+1/64)\\ &=c\ exp(-(x+1/8)(x+1/8)+1/64)\\ &=c\ exp(-(x+1/8)(x+1/8))*exp(1/64)\\ &=c*exp(1/64)exp\big(-(x+1/8)(x+1/8)\big) \end{align} $$

but I keep getting stuck. Does anyone have any ideas?

I'm given a pdf in a different form and asked to identify the random variable. Also, there is a positive constant $c$ that i have to identify.

PDF: $f(x)= c*\exp(-x^2-x/4)\ \forall x\in\mathbb{R} $

I suspect this is the normal distribution but I am not sure.

I have tried to factor the exponent by adding and subtracting 1/64

$$ \begin{align} c\ \exp(-x^2-x/4)&=c\ \exp(-(x^2+x/4+1/64-1/64))\\ &=c\ \exp(-(x^2+x/4+1/64)+1/64)\\ &=c\ \exp(-(x+1/8)(x+1/8)+1/64)\\ &=c\ \exp(-(x+1/8)(x+1/8))*\exp(1/64)\\ &=c\exp(1/64)\exp\big(-(x+1/8)(x+1/8)\big) \end{align} $$

but I keep getting stuck. Does anyone have any ideas?

I have a question that I am working on and am stuck. Could anyone lend their assistance?

I'm given a pdf in a different form and asked to identify the random variable. Also, there is a positive constant 'c'$c$ that i have to identify.

Here is the pdf

$ f(x)= c*exp(-x^2-x/4)$

$-infinity<x<infinity $ PDF: $f(x)= c*\exp(-x^2-x/4)\ \forall -\infty<x<\infty $

I suspect this is the normal distribution but I am not sure.

I have tried to factor the exponent by adding and subtracting 1/64

$cexp-(x^2+x/4)=cexp(-(x^2+x/4+1/64-1/64))=cexp(-(x^2+x/4+1/64)+1/64)=cexp(-(x+1/8)(x+1/8)+1/64)=c*exp(-(x+1/8)(x+1/8))*exp(1/64)$

$=(c*exp(1/64)(cexp(-(x+1/8)(x+1/8)$$$ \begin{align} c\ exp(-x^2-x/4)&=c\ exp(-(x^2+x/4+1/64-1/64))\\ &=c\ exp(-(x^2+x/4+1/64)+1/64)\\ &=c\ exp(-(x+1/8)(x+1/8)+1/64)\\ &=c\ exp(-(x+1/8)(x+1/8))*exp(1/64)\\ &=c*exp(1/64)exp\big(-(x+1/8)(x+1/8)\big) \end{align} $$

but I keep getting stuck. Does anyone have any ideas?

I have a question that I am working on and am stuck. Could anyone lend their assistance?

I'm given a pdf in a different form and asked to identify the random variable. Also, there is a positive constant 'c' that i have to identify.

Here is the pdf

$ f(x)= c*exp(-x^2-x/4)$

$-infinity<x<infinity $

I suspect this is the normal distribution but I am not sure.

I have tried to factor the exponent by adding and subtracting 1/64

$cexp-(x^2+x/4)=cexp(-(x^2+x/4+1/64-1/64))=cexp(-(x^2+x/4+1/64)+1/64)=cexp(-(x+1/8)(x+1/8)+1/64)=c*exp(-(x+1/8)(x+1/8))*exp(1/64)$

$=(c*exp(1/64)(cexp(-(x+1/8)(x+1/8)$

but I keep getting stuck. Does anyone have any ideas?

I'm given a pdf in a different form and asked to identify the random variable. Also, there is a positive constant $c$ that i have to identify.

PDF: $f(x)= c*\exp(-x^2-x/4)\ \forall -\infty<x<\infty $

I suspect this is the normal distribution but I am not sure.

I have tried to factor the exponent by adding and subtracting 1/64

$$ \begin{align} c\ exp(-x^2-x/4)&=c\ exp(-(x^2+x/4+1/64-1/64))\\ &=c\ exp(-(x^2+x/4+1/64)+1/64)\\ &=c\ exp(-(x+1/8)(x+1/8)+1/64)\\ &=c\ exp(-(x+1/8)(x+1/8))*exp(1/64)\\ &=c*exp(1/64)exp\big(-(x+1/8)(x+1/8)\big) \end{align} $$

but I keep getting stuck. Does anyone have any ideas?

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Can anyone identify this distribution?

I have a question that I am working on and am stuck. Could anyone lend their assistance?

I'm given a pdf in a different form and asked to identify the random variable. Also, there is a positive constant 'c' that i have to identify.

Here is the pdf

$ f(x)= c*exp(-x^2-x/4)$

$-infinity<x<infinity $

I suspect this is the normal distribution but I am not sure.

I have tried to factor the exponent by adding and subtracting 1/64

$cexp-(x^2+x/4)=cexp(-(x^2+x/4+1/64-1/64))=cexp(-(x^2+x/4+1/64)+1/64)=cexp(-(x+1/8)(x+1/8)+1/64)=c*exp(-(x+1/8)(x+1/8))*exp(1/64)$

$=(c*exp(1/64)(cexp(-(x+1/8)(x+1/8)$

but I keep getting stuck. Does anyone have any ideas?