How do I calculate z scores?

Apologies for the really basic question but I've never done probability before and am struggling with chapter 15.

How do I find the probability of a Z score?

For example, at the end of Q.4.6 in the Q&A bank it reads:

= P(Z > 1.165)
= 1-0.87799
= 0.12201

How did they arrive at that answer? Is it from the tables somewhere?

 

Hi,

The tables ( pg 160 and 161 of the Formulae and Tables book) give values of probabilities of x greater than Z,i.e P(Z<x). Here are certain formulas that may help you:-

1) if P(Z< -x)= P(Z>x)=1-P(Z<x)
2) if P(Z> -x)=P(Z<x).


In your example we have to find P(Z>1.165). So using the first formula it becomes-

1-P(Z<1.165).

The tables provide probabilities up to 3 significant figures,so we need to interpolate the probabilities. Now, we know that 1.165 lies between 1.16 and 1.17. the formula for interpolation is-

P(Z<1.16)+0.5{P(Z<1.17) - P(Z<1.16)}

(The 0.5 above is from the third significant figure. Suppose if it was 1.1669 instead of 1.165, then we need to put 0.69.)

From the tables we get:-

=0.87698+0.5{0.879-0.87698}
=0.87799

Therefore , 1-P(Z<1.165)=1-0.87799=0.12201

I really hope this helped!

Kid regards,
Sanjay.

 

Thanks so much Sanjay, that explains it perfectly.

@John Lee: So in the exam would it be alright to leave the answer as

1-P(Z<1.165)?

 

Lol no! We are still required to calculate the probability. We can either-

1) Calculate 1-P[Z<1.17](rounding 1.165 to 3 significant figures) -This involves looking up values of [Z<1.17] in the tables; 1-P[Z<1.17]=1-0.87900=0.121
or
2) Calculate 1-P[Z<1.165] - This requires us to interpolate the probabilities to find the value of [Z<1.165] which would give us a value of 0.12201.

What Sir meant is that the latter is not really required for CT 1, but is necessary for CT 3 and CT 6. If you look at the Examiner's report of the past papers in CT 1, you'll see that they have used interpolation. So I think that there's no harm in interpolating to just get the hang of it.

John Sir, thoughts?