S2 - Continuous Random Variables

Hi,

I would really appreciate some help with these questions:

1) I'm stuck on part c) of this question:

The amount of vegetables eaten by a family in a week is a random variable Wkg. The probability density function is given by:

f(w) = [20/(5^5)](w^3)(5-w) 0<=w<=5
0 otherwise

a) Find the cumulative distribution function of W.
which is:

F(w) = 0 x<0
[(w^4)/(5^5)](25-4w) o<=w<=5
1 w>5

b) Find to 3dp the probability that the family eats between 2kg and 4kg of vegetables in one week: 0.650

c) Verify that the amound, m, of vegetables such that the family is equally likely to eat more or less than m in any week is about 3.431kg.
How do you do this question?

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2) A continuous random variable X has probability density function, f, defined by:

f(x) = 1/4 0<=x<=1
(x^3)/5 1<=x<=2
0 otherwise

Obtain the cumulative distribution function and hence, or otherwise find to 3dp, the median and the interquartile range of the distribution.

I'm getting the CFD to be:

F(X) = 0 x<0
x/4 0<=x<1
1/4 + (x^4)/20 1<=x<=2
1 x>2

However, the answer in the book is given as:

F(X) = 0 x<0
x/4 0<=x<=1
1/5 + (x^4)/20 1<=x<=2
1 x>2

Where am I going wrong? To get F(X)= 1/4 + (x^4)/20 for 1<=x<=2 I integrated (x^3)/5 and then added on 1/4 from the previous line by substituting 1 into x/4.

Thank you very much for your help!

Reply 1

Trangulor

c) find the median. F(median) = 0.5

To get the CDF, all of it must add up to 1. Between 0 and 1, the probability is 1/4 (length times height). Hence the area between 1 and 2 must be 3/4

So between 1 and 2, you get (x^4)/20 + C
Using upper limit (2), (16/20) + C = 3/4, so C = 4/20 or 1/5

Your error is in taking taking the constant to be 1/4, presumably because that is what you should add on to the area between 1 and 2. You can do it this way (I'll show below), but since (1^4)/20 already has a value, you need to subtract this from 1/4

Using lower limit, your area should be 1/4:
1/4 = 1/20 + C so C = 1/5 again

Reply 2

sweet_gurl

Thanks for your help, I got question 2 done. I'm still a bit stuck on question 1) part c), this is how far I've got:

F(m)=1/2

(w^4)/125 -(4w^5)/3125 = 1/2
50w^4 - 8w^5 = 3125
50w^4 - 8w^5 - 3125 = 0

But where do I go from here to get w=3.431?

Thanks! :smile:

Reply 3

Trangulor

Since it says verify, I think you are allowed to simply work out F(3.431) by putting 3.431 into the equation, ie set w as 3.431 in the equation:

(w^4)/125 -(4w^5)/3125

and it should give you a half :smile:

Reply 4

sweet_gurl

Oh thats alright then, as that equation is horrible to solve. Thanks! :biggrin:

Reply 5

sweet_gurl

Can someone please check if this is correct (for another question)

The pdf for a continuous random variable:

f(x) = k(x^3) 0<=x<=1
(k/3)(3-x) 1<x<=3
0 otherwise

Find the value of k

I did ∫ (from 1 to 0) k(x^3) + ∫ (from 3 to 1) k - (kx)/3 dx = 1

[(kx^4)/4] (from 1 to 0) + [kx - (kx^2)/6)] (from 3 to 1) = 1

[(k/4)-0] + [(3k-3k/2)-(k-k/6)] = 1

11k/12 = 1

so k=12/11

Is this correct? I just thought the answer was a bit weird. Thanks!

Reply 6

dijon

Looks right to me

Reply 7

sweet_gurl

Trangulor

I tried doing it a different way, by putting

(w^4)/125 -(4w^5)/3125 - 1/2 =0

and then substituting the upper and lower bounds of 3.431 to find a change of sign. However, this didn't work, as when I put in 3.41315 I got
-0.00720324.... and when I put in 3.41305 I got -0.0072436... Do you know why this didn't work? I just thought it might be a better way of showing the answer is 3.431, or is it still best to put 3.431 into (w^4)/125 -(4w^5)/3125 and produce approximately 1/2?