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S2 pdf to cdf question help.

I am not sure on 7b as using the usual method where we put F(n)=0 or F(n)=1 doesn't work. I'm not sure how do we know in this question to put F(3)=1/5 and not F(4)=1/5 for the interval 3<x<4? I understand everything else but I'm not sure on that.
Thanks for the help.

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Reply 1

WhiteGroupMaths

You must be clear about this:

f(3) describes the specific value at the instance of the pdf curve when x=3, whilst F(3) refers to the aggregation of all probabilities(areas under relevant curve sections) from x= -∞ all the way up to x=3.

I shall get you started with the front end of things.

For 0 ≤x≤ 3, F(x)= ∫ x² / 45 dx = x³/135 (lower limit =0, upper limit =x)

So F(3)= P(X ≤ 3) = 3³/135 = 27/135 = 1/5

For the next segment, ie 3< x <4 ,

the pdf curve is simply described by a horizontal line.

Say we wish to calculate F(3.5), how can we go about doing so?

In this region, F(x)= (entire area under first curve for 0 ≤x≤ 3) + (area under second curve from 3 onwards to x)

=1/5 + ∫ 1/5 dx (lower limit of this integral=3, upper limit=x)

= 1/5 + 1/5( x-3) = 1/5x - 2/5

Then F(3.5) = P(X ≤ 3.5) = 1/5 * 3.5 -2/5 = 0.3

Hopefully your doubts are substantially cleared by now, and you can proceed to develop F(x) expressions for remaining intervals of x.

Peace.

(edited 9 years ago)