In: Statistics and Probability
A humanities professor assigns letter grades on a test according to the following scheme.
A: Top 8% of scores
B: Scores below the top 8% and above the bottom 58%
C: Scores below the top 42% and above the bottom 22%
D: Scores below the top 78% and above the bottom 7%
F: Bottom 7% of scores
Scores on the test are normally distributed with a mean of 77.1 and a standard deviation of 7.4.
Find the minimum score required for an A grade. Round your answer to the nearest whole number, if necessary.
Here we have = 77.1, = 7.4
A: Top 8% of scores
p ( Z > z ) = 0.08
p ( Z > z ) = 1- p ( Z z ) = 0.08
p ( Z z ) = 0.92
( We use excel formula "=norm.s.inv(0.92)" )
z = 1.41
x = 87.53
Minimum score required for an A grade is 87.53.
B: Below the top 8% and above the bottom 58%:
Here we need to find values a and b.
P ( Z < z ) = 0.58
( We use excel formula "=norm.s.inv(0.58)" )
We get z = 0.20
xa =78.58
And
p ( Z > z ) = 0.08
p ( Z > z ) = 1 - p ( Z z ) = 0.08
p ( Z z ) = 1 - 0.08 = 0.92
( We use excel formula "=norm.s.inv(0.92)" )
z = 1.41
xb = 87.53
C: Scores below the top 42% and above the bottom 22%
Here we need to find values a and b.
P ( Z < z ) = 0.22
( We use excel formula "=norm.s.inv(0.22)" )
We get z = -0.77
xa =71.40
And
p ( Z > z ) = 0.42
p ( Z > z ) = 1 - p ( Z z ) = 0.42
p ( Z z ) = 1 - 0.42 = 0.58
( We use excel formula "=norm.s.inv(0.58)" )
z = 0.20
xb = 78.58
D: Scores below the top 78% and above the bottom 7%
Here we need to find values a and b.
P ( Z < z ) = 0.07
( We use excel formula "=norm.s.inv(0.07)" )
We get z = -1.48
xa =66.15
And
p ( Z > z ) = 0.78
p ( Z > z ) = 1 - p ( Z z ) = 0.78
p ( Z z ) = 1 - 0.78 = 0.22
( We use excel formula "=norm.s.inv(0.22)" )
z = -0.77
xb = 71.40
F: Bottom 7% of scores
Here we need to find values a.
P ( Z < z ) = 0.07
( We use excel formula "=norm.s.inv(0.07)" )
We get z = -1.48
xa =66.15