Final averages are typically approximately normally distributed with a mean of 66 and a standard deviation of 10.5. Your professor says that the top 5% of the class will receive an A; the next 15%, a B; the next 47%, a C; the next 22%, a D; and the bottom 11%, an F.
(a) What average must you exceed to obtain an A?
(b) What average must you exceed to receive a grade of C or better?
(c) What average must you obtain to pass the course? (You'll need a D or better.)
I will let you play with
http://davidmlane.com/normal.html
click on "value from an area" and enter
mean: 66
sd: 10.5
click on below and enter .95 in area
I get 83.275
So 95% of the students will have a score of less than 83
the next 15 % will get a B
so 20 % will get either A or B
and 80% will not get A or B
click on below and enter .8 in area
I got 74.835 or 75% to 83 will get you a B
67% will get at least a C
enter .67 in area to get 70.615
I will let you continue
To answer these questions, we need to find the corresponding cutoff scores from the normal distribution based on the given class distribution. We can use the z-score formula to convert the cutoff percentiles to their corresponding z-scores. The z-score formula is given as:
z = (x - μ) / σ
where:
- z is the z-score
- x is the value we want to find the z-score for
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
(a) To obtain an A, we need to find the cutoff score for the top 5% of the class. To do this, we need to find the z-score corresponding to the 95th percentile.
Step 1: Find the z-score corresponding to the 95th percentile:
Given that the average (μ) is 66 and the standard deviation (σ) is 10.5, we can calculate the z-score using the formula:
z = (x - μ) / σ
The 95th percentile corresponds to a z-score of 1.645 (you can find this value using a standard normal distribution table or a calculator).
Step 2: Solve for x:
Rearranging the z-score formula, we get:
x = z * σ + μ
Substituting the values, we get:
x = 1.645 * 10.5 + 66
Simplifying the equation, we find that you must exceed an average of approximately 83.42 to obtain an A.
(b) To receive a grade of C or better, we need to find the cutoff score for the top 69% of the class (15% for B + 47% for C + 7% for D).
Step 1: Find the z-score corresponding to the 31st percentile:
The 31st percentile corresponds to a z-score of -0.483 (you can find this value using a standard normal distribution table or a calculator).
Step 2: Solve for x:
Using the z-score formula, we get:
x = z * σ + μ
Substituting the values, we get:
x = -0.483 * 10.5 + 66
Simplifying the equation, we find that you must exceed an average of approximately 61.86 to receive a grade of C or better.
(c) To pass the course, we need to find the cutoff score for the top 80% of the class (15% for B + 47% for C + 22% for D).
Step 1: Find the z-score corresponding to the 20th percentile:
The 20th percentile corresponds to a z-score of -0.841 (you can find this value using a standard normal distribution table or a calculator).
Step 2: Solve for x:
Using the z-score formula, we get:
x = z * σ + μ
Substituting the values, we get:
x = -0.841 * 10.5 + 66
Simplifying the equation, we find that you must obtain an average of approximately 57.39 to pass the course (receive a grade of D or better).