Joshua L. answered 05/09/24
Hi Xavier,
This problem involves the classic equation in introductory statistics:
z = (x - mu) / sigma
x= desired score
mu= mean
sigma= standard deviation
So:
x= 71
mu= 96
sigma= 19
z = (71 - 96) / 19
z = -1.32
From z-table:
P(Z< -1.32) = 0.0934
This is quite low, and it would make sense given how well students performed on that test on average. I hope this helps.