Urgent Query - Chances!!!

[jonbon]

Hi Everyone!

Could you kindly help me please. This has confused me forever!

For a standard normal distribution with

x p(x)
0 0.5
0.33 0.6293
0.67 0.74857
1 0.84135
1.33 0.90824
1.67 0.95254
2 0.97725

how is
1) z=-1 a 5 in a 6 chance?
2) z=4/3 a 1 in 11 chance
3) z=2/3 a 1 in 4 chance?

Is there a table of this chances?

Many thanks!

Jon.

 

[didster]

Where did you get that from?
I assume that you mean > rather than =.

10 in 11 = 0.909 ~ 0.908
etc.

It's sometimes easier for the audience to grasp fractions like this, so sometimes you find a fraction close to the real number and say almost 1 in 11.

 

[jonbon]

i know that they're approximately equal to the numbers I have provided. But in various CA3 examiner reports, these are mentioned as fractions! i want to know how to get to fractions? is it a case of memorising them? is there a list somewhere

 

[didster]

I doubt that there is a list, but it is becoming more and more apparent to me that actuarial exams are about lists so who knows.

You are probably familiar with the basic ones, eg halves, thirds, quarters, fifths, sixths, eighths, ninths, elevenths and twelfths. If you see a number close to one of these that you recognise, there you go. Also if 1/x = y, it's straightforward to find x.

Anything much more complicated than that may be difficult to spot and understood by the audience. Eg 5 in 63 or 42 in 1143 doesn't really help much.

I wouldn't worry about it too much, and if you're struggling, just quote the number or better yet a percentage.

 

[jonbon]

thanks.