Answer:
a) 7.68%
b) 3.38%
Explanation:
(a) Estimate the percentage of students scoring over 700 in 1967.
The probability that a student scored over 700 in 1967 equals the area under the Normal curve with mean 543 and standard deviation 110 to the right of 700.
In Excel this value is found with the formula
=1-NORMDIST(700,543,110,1)
and in OpenOffice Calc
=1-NORMDIST(700;543;110;1)
(NORMDIST(700;543;110;1) gives the area to the left of 700, so 1-NORMDIST(700;543;110;1) gives the area to the right of 700)
and equals 0.07675
So, the percentage of students scoring over 700 in 1967 was 7.68%
(b) Estimate the percentage of students scoring over 700 in 1994.
The probability that a student scored over 700 in 1994 equals the area under the Normal curve with mean 499 and standard deviation 110 to the right of 700.
In Excel this value is found with the formula
=1-NORMDIST(700,499,110,1)
and in OpenOffice Calc
=1-NORMDIST(700;499;110;1)
and equals 0.03383
So, the percentage of students scoring over 700 in 1994 was 3.38%