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If $X \sim \mathcal N(2,4)$, find the value of $P(X\ge3.5)$.

My attempt:

Given that, $\mu = 2$ and $\sigma^2 = 4$.

$\Bbb P(X \ge 3.5) = \int_{3.5}^x{\frac{1}{\sqrt{2 \cdot\pi \cdot4}}\cdot e^{\frac{-(t-2)^2}{2 \cdot 4}}}dt$

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    $\begingroup$ Why are you asking effectively the same question you asked yesterday? stats.stackexchange.com/questions/245582/… $\endgroup$ – mark999 Nov 14 '16 at 2:39
  • $\begingroup$ @mark999, coz, I wanted a different point of view, a different way of solving this problem. $\endgroup$ – user366312 Nov 14 '16 at 2:40
  • $\begingroup$ I'm probably wrong, but, for some reason, I thought N(mu,sigma) means the second number is the standard deviation, not the variance. $\endgroup$ – user1566 Nov 14 '16 at 3:09
  • $\begingroup$ @mark999, besides, there were no satisfactory answers. $\endgroup$ – user366312 Nov 14 '16 at 3:19
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Take a look at one of the definitions of the PDF,

$1 = \int_{-\infty}^{\infty}f(x)dx$

If f(x) has a support then,

$1 = \int_{S}f(x)dx$

Or like what @nafizh pointed out you could flip the integration or in it's current form use the limit from 3.5 to $\infty$. That's what I was hinting at.

The $x$ value would give you something similar to the CDF. The definition of the CDF is:

$ F_x(x) = \int_{-\infty}^{x}f(x)dx$

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You should calculate this by first changing the expression-

$P(X \ge 3.5) = 1 - P(X \lt 3.5)$

Then when you do the integration of the changed term, you have a limit of $-\infty$ to 3.5, which you can calculate easily.

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  • $\begingroup$ Not -infinity to 3.5 ? $\endgroup$ – user1566 Nov 14 '16 at 3:08

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