Denote the (100p) th percentile of the standard normal distribution as n(p).
It is known that if the (100p) th of the distribution of a random variable is n(p), then, \(P(X\leq n(p)) = p\), that is, \(\phi(n(p)) = p\).
\(100p = 91\)
Thus, the corresponding percentile is \(\phi (n(0.91)) = 0.91\).
In Table A-3 “Standard Normal Curve Areas” find the value of the variable which corresponds to the probability 0.9100.
Locate the probability value 0.9100 in the body of the table.
The exact value 0.9100 is not available. However, the probability 0.9099 is available, which is pretty close to 0.9100. That is, \(0.9100 \sim 0.9099\).
Move horizontally from the probability value till the first column is reached and note the value of z as 1.3.
Move vertically from the probability value till the first row is reached and note the value of z as .04.
\(\phi(1.34) = 0.9099\)
Now, \(\phi(n(0.91)) = 0.91\). Hence, the required percentile for the standard normal distribution is approximately 1.34.