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Suppose that a random variable $Y$ has a probability density function given by

$$f_Y(y) = \begin{cases} k y^3 e^{-y/2} & y > 0, \\ 0 & {\rm otherwise.} \end{cases}$$

a. Find the value of $k$ that makes $f_Y(y)$ a density function.

d. What is the probability that Y lies within 2 standard deviations of its mean?

My approach:

a. $$F(\infty) = \int_0^\infty k y^3 e^{-y/2} \, dy = 1.$$ I'm assuming if I carry out the rest of the integration and solve for $k$, this would make $f(y)$ a density function. Is this correct?

d. I am not exactly sure how to start d. Any suggestions?

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  • $\begingroup$ Yes, that's the correct start for $(a)$. $\endgroup$
    – user61527
    Feb 14, 2014 at 3:10
  • $\begingroup$ Great! Any ideas for d? $\endgroup$
    – Jebediah
    Feb 14, 2014 at 3:12
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    $\begingroup$ Write down (and evaluate) integrals for the mean $\mu$ and standard deviation $\sigma$. Then evaluate $$\int_{\mu - 2\sigma}^{\mu + 2\sigma} f(y) dy$$ $\endgroup$
    – user61527
    Feb 14, 2014 at 3:14
  • $\begingroup$ @Jebediah: Calculate the mean E(X), the standard deviation E(x-E(X)), etc. $\endgroup$
    – user99680
    Feb 14, 2014 at 3:14
  • $\begingroup$ @user99680 How does that help? $\endgroup$
    – Jebediah
    Feb 14, 2014 at 3:34

1 Answer 1

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In the same way as "a.", we can calculate $\mu=E[Y]$ and $E[Y^2]$.

So,we obtain \begin{eqnarray*} V[Y] &=& E[Y^2] - (E[Y])^2 \\ 2\sigma&=&2\sqrt{V[Y]} \end{eqnarray*} Then,we can calculate $$\int_{\mu - 2\sigma}^{\mu + 2\sigma} f(y) dy$$

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