Question

On your last exam, the class scored an average of 82 with a standard deviation of 8. What's the probability of any one student's score being 90 or greater?

Answer 1

solution:

Given that

Mean = = 82 Standard deviation = = 8

P( Any one students scoring being 90 or greater) = P(X >=90)

= P( X-/ >= x-/ )

=  P( X-/ >= 90-82/ 8 )

= P(Z>=1)

= 0.5 - P(Z=1)

= 0.5 - 0.3413 [ using Normal Distribution table]

= 0.1587

The probability of any one student's score being 90 or greater = 0.1587

The graph of P(X>=90) will be