Normal probability density function

Master J
Messages
219
Reaction score
0

Homework Statement



A production line is producing cans of soda where the volume
of soda in each can produced can be thought of as (approximately) obeying a normal distribution
with mean 500ml and standard deviation 0.5ml. What percentage of the cans will have more than
499ml in them?


Now at my level, the functions to use were simple given to us. The normal probability density function and the standard density function. However, I don't quite understand the idea of P(>499) = 1 - P(<499) = 1 - F(499).

Could anyone explain why that is?

Thanks
 
Physics news on Phys.org
I'm not 100% sure of your question, so if this doesn't answer it try again.

If F is a cumulative distribution function (normal or any other) for a continuous random variable, then for any number x, this is true:

<br /> F(x) = \Pr(X \le x )<br />

This means that whenever you need a probability that is of the form \Pr(X &gt; x), you need to rewrite this in terms of the complement inequality:

<br /> \Pr(X &gt; x) = 1 - \Pr(X \le x)<br />

This is what happens in your problem, with x = 499. Is this the type of thing you want?
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top