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Edexcel S2 Ex.2D Q11

I need to know if this is the correct approach to this question, or is there a better method.

A continuous random variable has p.d.f. f, defined by

..........¼, 0<=x<1
f(x) = x³/5 1<=x<=2
..........0 otherwise


Obtain the c.d.f. (the median and IQR are to be found too, but I can do that ok.)

The pdf has a "step" down above x= 1, where f(x) = ¼ and 1/5

The cdf must be a smooth line (no "steps")(Is that right?)
Integrating f(x)

..........0 x<0
..........x/4, 0<=x<1
f(x) = (x^4)/20 +C 1<=x<=2
..........0 otherwise

find C. At x=1, ¼ = 1/20 +C, so C = 1/5 (I think it's correct that at no point can the cdf gradient be negative, so a step down is not possible.)

I could use x=2, so that 16/20 +C =1
Is this all correct?

Fancy another? My thread about q9 of the same exercise has so far attracted slightly less than one reply...
http://www.thestudentroom.co.uk/t195185.html

Aitch
Hey :smile:
I just looked throught my book and all ur working here seems right
On this question, when finding the median etc, do u just put the x^4/20 +1/5 = 1/2 ....... what about the x/4 bit of the function?
you just ignore the x/4 bit because when you put in F(1) (the maximum limit for that bit), you get 1/4 which is lower than 0.5. this means the median cannot lie in that particular part of the function. It therefore must lie in the other part. so m^4/20 + 1/5 = 1/2 is correct.

I hope that made sense
Yeah it did, thanks :smile:
Reply 5
Avatar for Aitch
Aitch
OP
2
Featherflare
you just ignore the x/4 bit because when you put in F(1) (the maximum limit for that bit), you get 1/4 which is lower than 0.5. this means the median cannot lie in that particular part of the function. It therefore must lie in the other part. so m^4/20 + 1/5 = 1/2 is correct.

I hope that made sense


Yes, it did. This was how I approached that bit. The IQR is quite easy too, since q1 = 1
Reply 6
Avatar for samd
samd
10
An alternatice method is as follows:

Finding CDF

For 0 < x < 1: F(x0x_{0}) = ∫0x014dx=x4\int^{x_0}_{0}\frac{1}{4}dx = \frac{x}{4}

For 1 < x 2: F(x0x_{0}) = ∫0114dx+∫1x0x35dx=x0420+15\int^{1}_{0}\frac{1}{4}dx + \int^{x_0}_{1}\frac{x^{3}}{5}dx = \frac{x_{0}^{4}}{20} + \frac{1}{5}

So in this method you just integrate every interval that comes before the ones you are interested in across it's full limits, and add it on.
Reply 7
Avatar for Aitch
Aitch
OP
2
samd
An alternatice method is as follows:

Finding CDF

For 0 < x < 1: F(x0x_{0}) = ∫0x014dx=x4\int^{x_0}_{0}\frac{1}{4}dx = \frac{x}{4}

For 1 < x 2: F(x0x_{0}) = ∫0114dx+∫1x0x35dx=x0420+15\int^{1}_{0}\frac{1}{4}dx + \int^{x_0}_{1}\frac{x^{3}}{5}dx = \frac{x_{0}^{4}}{20} + \frac{1}{5}

So in this method you just integrate every interval that comes before the ones you are interested in across it's full limits, and add it on.


Thanks for the alternative method.

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