How do I standardise scores?

Question

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

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Answer 1 · 2018-07-06T07:47:45

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#