Question

How do I standardise scores?

Mean = 90
Standard dev= 2
P(87.2<X<k) = 0.5964
I need to find k.

Answer 1

Usually normal tables distribution are N(0,1). This is the reason because we normalize. In our case we have

Normalize is equivalent to make a variable change in form

`Z=(X-mu)/sigma` where `X` is the given variable, `Z` is normalized variable and `mu, sigma` are mean and standard deviation.

In our case: `P(87.2< X< K)=P((87.2-90)/2 < Z <(K-90)/2)=`

`=P(-1.4< X <(K-90)/2)=P(X < (K-90)/2)-1+P(X<1.4)`

Depending on Normal table available these values will be calculated in one way or in other but with same results...

In my case I have a Normal table which gives `P(X<=a)`. Then

`P(-1.4 < X)=1-P(X<1.4)=1-0.9192=0.0808`. So, we have to calculate

`0.5964=P(X < (K-90)/2)-0.0808`

`P(X < (K-90)/2)=0.6772`. Looking at tables again

`(K-90)/2=0.46` transposing terms

`K=90.92`