# Thread: [SOLVED] Random Variables -Continuous Prob Distributions

1. ## [SOLVED] Random Variables -Continuous Prob Distributions

Qn1 )
An important factor in solid missile fuel is the particle size distribution. Significant probs occur if the particle sizes are too large. From production data in the past, it has been determined that the particle size (in micrometers) distribution is characterised by :

f(x) = 3x (to the power of -1/4) , x>1,
0, elsewhere

a) Verify that this is a valid density function.
b) Evaluate F(x)
c) What is the probability that a random particle from the manufactured fuel exceeds 4 micrometers?

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a) How do I verify ?
b) I integrate 3x (to the power of -4) from 0 to 1 to dx.
but it is wrong.
[if i got a) and b) , i shld be able to do c) which I think is P (x>4) ]

Thks.

2. x>1, so you integrate from 1 to infinity
This is a pareto rv.

-1/4 is wrong, is the power -4?

$\int_1^{\infty} 3x^{-4}dx=1$

$F(x)= \int_1^x f(t)dt=\int_1^x 3t^{-4}dt=1-x^{-3}$ where x>1

and 0 otherwise

3. Originally Posted by matheagle
x>1, so you integrate from 1 to infinity
This is a pareto rv.

-1/4 is wrong, is the power -4?

$\int_1^{\infty}3x^{-4}dx=1$

$F(x)=\int_1^x f(t)dt$ where x>1

oh yes. this is ^-4 and not ^-1/4.
from walpole 3.29.

so in order to verify, integrate 3x^-4,
= [-x^-3] from 1 to infinity,
and since infinity to power ^-3 is near 0 ,
= 0 - (-1)^-3
=1

am i right ?

but how should i evaluate F(x) ?
this is confusing.

4. almost

it's = 0 -- (1)^-3 = 0 + (1)^-3=1

YOU plug in 1 not -1

5. ok, got it . thks. let me try part c) now

6. the answer = 0.0156 which is 4^-3.

but shouldn't it be [1-x^-3 ] from 1 to 4 ?

7. ok i got the answer. my careless mistake.

,

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### An important factor in solid misile fuel is the particle size distribution

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