(New page: == Find the value of c such that == <math> f(x)= \begin{cases} \frac {x} {6} + c & 0\le x \le 3 \\ 0 & \mbox{elsewhere} \end{cases} </math> is a p.d.f. Also find <math>P(1\le x \l...) |
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− | == Find the value of c such that == | + | [[Category:ECE302Fall2008_ProfSanghavi]] |
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+ | == Question: Find the value of c such that == | ||
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<math>f(x)\le 0 if c\ge 0</math>; Also we must have | <math>f(x)\le 0 if c\ge 0</math>; Also we must have | ||
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<math>P(1\le x \le 2)= \frac {1} {3}</math> | <math>P(1\le x \le 2)= \frac {1} {3}</math> | ||
+ | ---- | ||
+ | [[Main_Page_ECE302Fall2008sanghavi|Back to ECE302 Fall 2008 Prof. Sanghavi]] |
Latest revision as of 13:29, 22 November 2011
Question: Find the value of c such that
$ f(x)= \begin{cases} \frac {x} {6} + c & 0\le x \le 3 \\ 0 & \mbox{elsewhere} \end{cases} $
is a p.d.f. Also find $ P(1\le x \le 2) $
Solution:
$ f(x)\le 0 if c\ge 0 $; Also we must have
$ \int\limits_{-infty}^{infty}f(x)dx=1 $
i.e., $ \int\limits_{0}^{3} ( \frac {x} {6} + c)dx=1 $
On integrating this we get,
$ \frac {3} {4} + 3c=1 $
Therefore,
$ c=\frac {1} {12} $
Now,
$ P(1\le x\le 2)=\int\limits_{1}^{2} f(x) dx $
$ =\int\limits_{1}^{2}(\frac {x} {6} + \frac {1} {12}) dx $
Which on further integration leads to,
$ \frac {1} {12} [(4+2)-(1+1)]=\frac {1} {3} $
Thus,
$ P(1\le x \le 2)= \frac {1} {3} $