We are looking for the probability :
[tex]P(X > n)=0.05[/tex]
Transform the law to standard normal like this:
[tex]P(\dfrac{X-545}{100}> \dfrac{n-545}{100})=0.05[/tex]
The above formula is equivalent to this one:
[tex]P(\dfrac{X-545}{100}< \dfrac{n-545}{100})=0.95[/tex]
From normal law table, we read the value of [tex]\dfrac{X-545}{100}[/tex].
[tex]\dfrac{X-545}{100}=1.7[/tex]
Solving the above equation for the score n:
[tex]X-545=170[/tex]
[tex]X=715[/tex], it is the score we are looking for.
