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.
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?
Z = (x - mean)/SD
Substitute known values to find x.
1. Find values in equation above for Z = ±1
2. Find Z scores in table (below) for both points between those points and the mean.
3. In normal distribution, mean = median = mode.
4. From table of "areas under normal distribution" in the back of your stats text, what Z score corresponds to .90 in larger area?
5. Compare your Z score with her Z score.
I hope this helps.
To find the scores of the students who were within one standard deviation of the mean, we need to calculate the range. In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean.
To find the lower and upper bounds, we can subtract and add one standard deviation from the mean:
Lower bound = mean - standard deviation = 195 - 30 = 165
Upper bound = mean + standard deviation = 195 + 30 = 225
So, the scores of the students who were within one standard deviation of the mean ranged from 165 to 225.
To calculate the percent of students that includes, we can use the concept of the Empirical Rule. The Empirical Rule states that in a normal distribution, approximately 68% of the data falls within one standard deviation of the mean.
Therefore, approximately 68% of students' scores were within one standard deviation of the mean.
To find the score of students who scored in the middle of the class (50% did better, 50% did worse), we can calculate the median. In a normal distribution, the median is equal to the mean.
So, the score of students who scored in the middle of the class was 195.
To find the minimum Z-score of someone getting an A (90% or above), we can use the Z-score formula. The Z-score measures how many standard deviations a data point is from the mean.
Z-score = (X - mean) / standard deviation
To get a minimum Z-score of someone getting an A (90% or above), we need to find the Z-score corresponding to a cumulative probability of 90%. This Z-score will give us the minimal score needed to be in the top 10%.
Using a Z-table or a statistical calculator, we can find that the Z-score for a cumulative probability of 90% is approximately 1.28.
So, the minimum Z-score of someone getting an A is 1.28.
To determine who did better between you and your colleague, we can calculate both of your Z-scores and compare them.
For you:
Z-score = (X - mean) / standard deviation = (235 - 195) / 30 = 1.33
For your colleague:
Z-score = (X - mean) / standard deviation = (82 - 72) / 7 = 1.43
Comparing the two Z-scores, your colleague has a higher Z-score of 1.43, indicating a better performance in her section of the course.