Continuous Probability functions

**AshleyT** · May 12th 2011, 03:08 PM

http://myasdcommunity.com/continuous.jpeg
(I keep getting not a valid image file, sorry, so couldn't display :S...)

I've managed the first two parts of the question. How ever i was a bit confused as for how the second question the solutions managed to obtain:

$1- \frac {1}{x^2}$

For the integral of the function.

Thus for the third part of the question, i am confused:
It would make sense to have to work out P(X >= 4) which would be
1 - P(X <= 3)

However they sub in 4 into an equation and get:
$1 - FT (4) = 1 - (1 - (1 - \frac{1}{4^2} )= 1/16$

And i am very confused as to where this comes from :S

Thank-you.

**mr fantastic** · May 12th 2011, 08:18 PM

Originally Posted by AshleyT

http://myasdcommunity.com/continuous.jpeg
(I keep getting not a valid image file, sorry, so couldn't display :S...)

I've managed the first two parts of the question. How ever i was a bit confused as for how the second question the solutions managed to obtain:

$1- \frac {1}{x^2}$

For the integral of the function.

Thus for the third part of the question, i am confused:
It would make sense to have to work out P(X >= 4) which would be
1 - P(X <= 3)

However they sub in 4 into an equation and get:
$1 - FT (4) = 1 - (1 - (1 - \frac{1}{4^2} )= 1/16$

And i am very confused as to where this comes from :S

Thank-you.

The solutions are using the cdf of X:

$cdf = F(x) = \int_1^x \frac{d}{t^3} \, dt = \frac{d}{2} \left(1 - \frac{1}{x^2}\right)$

Since $F(+\infty) = 1$ it's clear that d = 2.

So $F(x) = 1 - \frac{1}{x^2}$ .

Then you require $\Pr(X \geq 4) = 1 - \Pr(X < 4) = 1 - F(4)$ .

I suggest you review the definition, calculation and application of the cumulative distribution function cdf.

**AshleyT** · May 12th 2011, 11:51 PM

Originally Posted by mr fantastic

The solutions are using the cdf of X:

$cdf = F(x) = \int_1^x \frac{d}{t^3} \, dt = \frac{d}{2} \left(1 - \frac{1}{x^2}\right)$

Since $F(+\infty) = 1$ it's clear that d = 2.

So $F(x) = 1 - \frac{1}{x^2}$ .

Then you require $\Pr(X \geq 4) = 1 - \Pr(X < 4) = 1 - F(4)$ .

I suggest you review the definition, calculation and application of the cumulative distribution function cdf.

Thank-you.
I've been reading notes on everything, however because i have ADD, i have to learn by example and have been struggling to find any except for a few questions which had poorly written solutions with no explanations. When i was taught it 3 years ago, i was taught to do it differently which is why i was getting confused, assuming they were doing things the same way.

I must have made an error however as i was not achieving the same answer at 1/16 with my method.

Completely understand now, thankyou

.

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