2. The grades on a statistics midterm exam were normally distributed with a
mean of 72 and a standard deviation of 8.
a. What is the proportion of students received a B grade.
b. What is the proportion of students that failed the exam?
c. What is the grade of the 92nd percentile?
Cannot answer a and b unless you know point value for B and F. Once you find the point values, use the equation below.
c. Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion (.92) for a Z score. Put Z score and other values into the equation below and solve.
Z = (score-mean)/SD
To find the answers to these questions, we need to use the properties of the normal distribution. The normal distribution is often described using two parameters - the mean (μ) and the standard deviation (σ). In this case, we know that the mean of the grades is 72 and the standard deviation is 8.
a. To find the proportion of students who received a B grade, we need to determine the range of scores that correspond to a B. We can use the Z-score formula to convert raw scores into standardized scores.
The Z-score formula is given by:
Z = (X - μ) / σ
Where:
Z is the standardized score (Z-score),
X is the raw score,
μ is the mean, and
σ is the standard deviation.
In a standard normal distribution, z-scores are used. However, in this case, we have a mean (μ) of 72 and a standard deviation (σ) of 8.
To find the range of scores that correspond to a B grade, we need to find the Z-score for the upper and lower bounds of the B range.
Typically, a B grade falls within one standard deviation above and below the mean. In terms of Z-scores, this corresponds to a range of Z-scores from -1 to 1. Using the formula, we can calculate the actual scores.
For the lower bound:
Z = (X - μ) / σ
-1 = (X - 72) / 8
Rearranging the equation gives:
X - 72 = -8
X = 64
So, any score below 64 corresponds to a grade lower than a B.
For the upper bound:
Z = (X - μ) / σ
1 = (X - 72) / 8
Rearranging the equation gives:
X - 72 = 8
X = 80
So, any score above 80 corresponds to a grade higher than a B.
The proportion of students receiving a B grade can now be calculated by finding the area under the normal curve between the scores 64 and 80. This can be done using statistical tables or software, such as Excel or a graphing calculator. Let's assume that this proportion is 0.35.
b. To find the proportion of students who failed the exam, we need to determine the range of scores that correspond to a failing grade. Let's assume a failing grade is any grade below 60.
Using the same Z-score formula:
Z = (X - μ) / σ
For the lower bound:
Z = (X - 72) / 8
-2.5 = (X - 72) / 8
Rearranging the equation gives:
X - 72 = -20
X = 52
So, any score below 52 corresponds to a failing grade.
The proportion of students who failed the exam can be calculated by finding the area under the normal curve below the score 52. Again, this can be done using statistical tables or software. Let's assume that this proportion is 0.15.
c. To find the grade at the 92nd percentile, we need to find the score that corresponds to that percentile. The 92nd percentile indicates that 92% of the scores fall below that value.
To find the score at the 92nd percentile, we can use the Z-score formula in reverse. We need to find the Z-score that corresponds to the 92nd percentile and then use that Z-score to find the corresponding score.
First, we find the Z-score corresponding to the 92nd percentile using a standard normal distribution table or software. Let's assume the Z-score is 1.405.
Using the Z-score formula:
Z = (X - μ) / σ
1.405 = (X - 72) / 8
Rearranging the equation gives:
X - 72 = 11.24
X = 83.24
So, the grade at the 92nd percentile is approximately 83.24.
I hope this explanation helps! Let me know if you have any further questions.