A students grade on an examination was transformed to a z va

A student\'s grade on an examination was transformed to a z value of 0.67. Assuming a normal distribution, we know that she scored approximately in the top:

Solution

Let Z denote the z-values of the scores. Then, Z is distributed as a standard normal.

i.e., Z~N(0,1)

P(Z>=0.67)=1-P(Z<0.67)=1-0.7485711 = 0.2514289 = 0.25 (approx)

This means the fraction of students with z-value of their scores greater than or equal to 0.67 is approx 0.25.

Thus, the students with z-value of her score lies in top 25%.