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Find$C$. $\mu$ and $c^{2}$ of the normal distribution whose probability function is given by $f{x}$=$C$$e^{\large -x^2+3x}$$,-\infty<$X$<\infty$.

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Toolbox:
  • A Continuous random variable $X$ is said to follow a normal distribution with parameter $\mu$ and $\sigma$ (or $\mu$ and $\sigma^2$) if the probability density function is
  • $f(x)=\large\frac{-1}{\sigma \sqrt{2\pi}}$$e^{-\Large\frac{1}{2}(\Large\frac{x-\mu}{\sigma})^2};-\infty < x < \infty,-\infty < \mu < 0$ and $\sigma >0$
  • $X\sim N(\mu,\sigma).$
  • Constants of a normal distribution :
  • Mean $=\mu$
  • variance=$\sigma^2$
  • standard deviation=$\sigma$
Step 1:
The probability density function of a random variable $X$
$f(x)=Ce^{x^2-3x},-\infty < x < \infty$
Step 2:
$f(x)=Ce^{(-x^2-3x+\Large\frac{9}{4}-\frac{9}{4})}$
$\quad\quad=Ce^(-x^2-3x+\large\frac{9}{4})$$+\large\frac{9}{4}$
$\quad\quad=Ce^{\Large\frac{9}{4}}.e^{-(x-\Large\frac{3}{2})^2}$
$\quad\quad=Ce^{\Large\frac{9}{4}}.e^{-\large\frac{1}{2}\big(\Large\frac{x-\large\frac{3}{2}}{\Large\frac{1}{\sqrt 2}}\big)^2}$
Step 3:
Comparing with the probability density function $f(x)=\large\frac{-1}{\sigma \sqrt{2\pi}}$$e^{-\Large\frac{1}{2}(\Large\frac{x-\mu}{\sigma})^2}$
We can see that $\mu=\large\frac{3}{2}$$,\sigma=\large\frac{1}{\sqrt 2}$
$\large\frac{1}{\sigma \sqrt{2\pi}}=\frac{1}{\Large\frac{1}{\sqrt 2}.\sqrt 2\pi}$
$\qquad\;\;=\large\frac{1}{\sqrt{\pi}}$
$\qquad\;\;=Ce^{\Large\frac{9}{4}}$
$\large\frac{1}{\sqrt\pi}$$=Ce^{\Large\frac{9}{4}}$
$C=\large\frac{e^{-2.25}}{\sqrt{\pi}}$
answered Sep 20, 2013 by sreemathi.v
 

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