Students were given an exam with 300 multiple-choice questions. The distribution of the scores was normal and mean was 195 with a standard deviation of 30. You may find it helpful to draw out this distribution before answering the questions below
What were the scores of the students who were within one standard deviation of the mean?
What percent of students did that include?
What was the score of students who scored in the middle of the class? (50% did better, 50% did worse).
If A’s are given to students who score 90% or above, what is the minimum Z-score of someone getting an A?
Let’s say you got 235 on this test. Your colleague is in a different section of this course. Her score was 82 out of 100 on her test. In her section the mean was 72 and the standard deviation was 7. Which one of you did better? (Hint: compare Z-scores.)
Try using the normal distribution tool at
http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html
68.27% of the students are within one standard deviation of the mean (165 to 225)
A 90% score would have been 270 correct answers.
Only 0.6% got that score or higher. The Z-score is 2.5 in that case. (2.5 standard deviations above the mean)
Repeat procedure for the second distribution and compare Z-scores
thank you
To answer these questions, we can use the properties of a normal distribution, specifically the mean and standard deviation. Let's break down each question and explain how to solve it:
1. What were the scores of the students who were within one standard deviation of the mean?
To find the scores within one standard deviation of the mean, we need to determine the range. Since the standard deviation is 30, we can calculate the scores by adding/subtracting 30 from the mean. So, the scores within one standard deviation of the mean would be between 195 - 30 = 165 and 195 + 30 = 225.
2. What percent of students did that include?
To calculate the percentage of students within one standard deviation of the mean, we can use the empirical rule for normal distributions. According to the empirical rule, approximately 68% of the data falls within one standard deviation of the mean. Therefore, approximately 68% of the students scored between 165 and 225.
3. What was the score of students who scored in the middle of the class (50% did better, 50% did worse)?
In a normal distribution, the median (50th percentile) is equal to the mean. Since the mean is given as 195, the score of students who scored in the middle of the class is 195.
4. If A’s are given to students who score 90% or above, what is the minimum Z-score of someone getting an A?
To find the minimum Z-score of someone getting an A, we need to convert the score of 90% into a Z-score using the formula:
Z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation. Rearranging the formula, we get:
X = Z * σ + μ.
Substituting the known values, we can calculate the minimum score as follows:
X = (90 * 30) + 195 = 282. Therefore, the minimum score to get an A is 282.
5. Which one of you did better? (Hint: compare Z-scores)
To compare the performance of you and your colleague, we need to standardize both scores using Z-scores. For you, with a score of 235, the Z-score can be calculated using the formula discussed earlier:
Z = (X - μ) / σ.
Substituting the given values:
Z = (235 - 195) / 30 = 1.33.
For your colleague, with a score of 82, we can calculate her Z-score using the same formula with the mean and standard deviation of her section:
Z = (82 - 72) / 7 = 1.43.
Comparing the Z-scores, we can see that a higher Z-score indicates a better performance. Therefore, your colleague, with a Z-score of 1.43, did better compared to you, with a Z-score of 1.33.