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As the title says, I got an exercise I don't know how to approach:

Let $X$ be a normal random variable, with Expected value of 12 and Variance of 4.

Find $C$ such that $P(X > C) = 0.1 $

How should I approach such question? First time I see the difinitions Expected value, variance.

Thanks in advance.

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  • $\begingroup$ This is the same as finding $C$ such that $P\{X \le C \} = 0.9$. $\endgroup$
    – copper.hat
    Dec 5, 2014 at 20:19
  • $\begingroup$ So what do you know about normal random variables (other than they are a kind of random variable that are normal instead of abnormal) $\endgroup$
    – Dilip Sarwate
    Dec 5, 2014 at 20:26

1 Answer 1

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$P(X>C)=0.1\leftrightarrow P((X-12)/2>(C-12)/2)=0.1$

$\rightarrow P(Z >(C-12)/2)=0.1$, where $Z = (X-12)/2$

Z~standard normal, Z is normal since it is a linear combination of normal random independent variables, this can be proven using moment-generating functions. And you should check that the expectation of $(X-12)/2$ is 0 and that its variance is 1, for this you use the general rules for calculating expected value and variance.

$\rightarrow P(Z\le(C-12)/2)=1-0.1=0.9$

$\Phi^{-1}(0.9)=1.281552$, from tables

$\rightarrow (C-12)/2=1.281552\rightarrow C=14.56$

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  • $\begingroup$ what do you mean by "linear combination of normal random variableS"? here there is just one variable. C is a constant: what do you mean by (C-12)/2 having a zero mean? what is $\Theta^{-1}$ etc $\endgroup$
    – Math-fun
    Dec 5, 2014 at 20:39
  • $\begingroup$ You can prove using moment generating functions that a linear combination of independent normal random variables+ a constant is also normally random variables, we have (X-12)/2=X/2-6, which is a linear combination of of 1 normal random variable and a constant. Even though in this case you really only need that if X is normal, then aX+b is also normal, where a and b are constants. $\endgroup$
    – user119615
    Dec 5, 2014 at 20:48
  • $\begingroup$ I meant to write X instead of C in (C-12)/2 that you meantion. The inverse function I posted is the inverse of the cumulative distribution function of the standard normal random variable: en.wikipedia.org/wiki/… $\endgroup$
    – user119615
    Dec 5, 2014 at 20:51

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