Answer:
a) 7.68%
b) 3.38%
Step-by-step 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<em> Excel </em> this value is found with the formula
=1-NORMDIST(700,543,110,1)
and in <em>OpenOffice Calc
</em>
=1-NORMDIST(700;543;110;1)
<em>
(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)
</em>
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 <em>Excel</em> this value is found with the formula
=1-NORMDIST(700,499,110,1)
and in <em>OpenOffice Calc
</em>
=1-NORMDIST(700;499;110;1)
and equals 0.03383
So, the percentage of students scoring over 700 in 1994 was 3.38%
