Answer:
Step-by-step explanation:
Given that the time to complete a standardized exam is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes.
P(completing exam before 1 hour)
= P(less than an hour) = P(X<60)
=P(Z<[tex]\frac{60-70}{10} =-1[/tex])
=0.5-0.34=0.16
i.e. 16% of students completed the standardized exam.
