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The density function of a random variable x is $f(x)=ke^{-2x^{2}+10x}$. Find the upper 5% point of the distribution of the means of the random sample of size 25 from the above population.

I need hints to pursue this. My thoughts:

  1. Assume normal population. Find the value of k by integrating $\displaystyle \int_{-\infty}^{\infty} ke^{-2x^{2}+10x}dx=1$

  2. Find $\mu$ by integrating $\displaystyle \int_{-\infty}^{\infty} xke^{-2x^{2}+10x}dx$

  3. Find $\sigma^2$ by integrating $\displaystyle \frac{1}{24}\int_{-\infty}^{\infty} (x-\mu)^2ke^{-2x^{2}+10x}dx$

  4. The upper 5% point of the distribution of the means of the random sample will correspond to $\displaystyle z_{0.95}=\frac{(x-\mu)}{\sigma}$.

Should $ z_{0.95}$ be $1.96$ (two-sided) or $1.645$(one-sided)?

Finally, replace values of $z_{0.95}$, $\mu$ and $\sigma$ to get the value of x, which is the required answer.

Are these steps correct?

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    $\begingroup$ You don't need to assume a population distribution; the density is given to you. You should be able to show that it's normal. I suggest you try completing the square in the exponent, it will save some effort. $\endgroup$
    – Glen_b
    Commented May 24, 2014 at 12:04
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    $\begingroup$ To continue @Glen_b's wise hint, you do not have to compute any integrals at all. That is because (1) this question evaluates your understanding of the relationship between the parameters (mean and SD) of a distribution and the parameters of its sampling distribution and (2) the parameters of the parent distribution are easily found by examining the $-2x^2+10x$ term alone: use algebra to rewrite it in the form $(x-\mu)^2/(2\sigma^2) + c$ where $c$ is not a function of $x$. $\endgroup$
    – whuber ♦
    Commented May 24, 2014 at 18:44
  • $\begingroup$ I did not understand .. and I am stuck anyways :( $\endgroup$
    – square_one
    Commented May 27, 2014 at 5:16

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The term in the exponent is $-2x^2+10x$.

I mentioned that this is done by completing the square. As whuber indicates, you need to write $-2x^2+10x$ in the form $-(x-\mu)^2/(2\sigma^2) + c$

$-2x^2+10x=-2(x^2-5x^2+(\frac52)^2)+25/2$

$\qquad\qquad\quad\:\,=-2(x-\frac52)^2+25/2$

$\qquad\qquad\quad\:\,=-\frac12\frac{(x-\frac52)^2}{\frac12^2}+25/2$

So we can immediately write down the mean and variance of the original normal distribution. This is what it means to "complete the square", though we don't usually need to spell it out in this level of detail.

The rest should be straightforward. (With the mean and variance of Gaussian $X$ you can work out the distribution of $\bar{X}$ and then the upper 5% point)

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  • $\begingroup$ +1. Given the OP exhibits a knowledge of Calculus, it might be informative to share a simpler solution method. Begin by recognizing this is a Normal PDF. To find the mean, one needs only to obtain the point where $f$ is largest. Because exponentiation is an increasing function, that will be where $-2x^2+10x$ is greatest, whence its derivative $-4x+10$ will be zero there. The unique solution $x=10/4=5/2$ therefore is the mean. The variance will be the reciprocal of the coefficient of $-x^2/2$, so by noting $-2x^2=4(-x^2/2)$ we immediately obtain $1/4$ for the variance. $\endgroup$
    – whuber ♦
    Commented Jun 21, 2016 at 13:46
  • $\begingroup$ I also notice that the OP appeared to neglect the part about "random sample of size $25$", so that deserves a mention. Basic properties of expectation and variance imply such samples will have the same mean (ergo, $5/2$) and $1/25$ of the variance of the parent distribution. Thus, the correct calculation involves a Normal$(5/2, (1/4)/25)$ distribution. In the OP's notation, $\mu=5/2$ and $\sigma=\sqrt{(1/4)/25}=1/10$. Finally, there's obviously no question of "one-sided" or "two-sided": those concepts apply to testing, not to this textbook problem asking for the "upper 5% point." $\endgroup$
    – whuber ♦
    Commented Jun 21, 2016 at 13:50
  • $\begingroup$ @whuber I spelled out what "the rest should be straightforward" a little more, to clarify that it includes the part about the sample mean. $\endgroup$
    – Glen_b
    Commented Jun 21, 2016 at 14:10

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