S2 - here we go again... Watch

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-=Rg=-
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Report 17 years ago
#1
Hey - sorry to keep posting about S2... it'll be over soon so you need not worry about being
bombarded with questions beyond the end of the week!

Basically,

for a p.d.f.
f(x) = (1/9)(8-x) 5=< x < 8

now, when I integrate this:

(1/9) int (between 5 and x0) 8-x dx

I therefore get F(x) for this part of the f(x) to equal:

(1/10)(16x - x^2 - 55)

The book seems to think it is as above but with -46 instead of -55.

For those with the book "Revise for Stats 2" this question is taken from 7c) of the practice paaper
in the rear of the book.

Is the book wrong? Or is it me?

Hope someone can help.

Thanks

RG
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