The mean score on a Math exam was 50% and the standard deviation was 10%. Assume the results were normally distributed.
(a) What percent of students earned a score between 50% and 70%?
( /2)
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to Z score. Change to percentage.
To find the percent of students who earned a score between 50% and 70%, we need to first convert the scores from percentages to z-scores. The formula to convert a score to a z-score is:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the raw score
- μ is the mean
- σ is the standard deviation
In this case, the mean score is 50% and the standard deviation is 10%. So, the formula becomes:
z = (x - 50) / 10
Now, let's calculate the z-scores for the scores of 50% and 70%:
z1 = (50 - 50) / 10 = 0
z2 = (70 - 50) / 10 = 2
Next, we need to find the area under the normal distribution curve between these two z-scores, which represents the percentage of students who earned scores between 50% and 70%.
To find this area, we can use a standard normal distribution table or a statistical calculator.
Using a standard normal distribution table, we can look up the z-scores of 0 and 2. The area between these two z-scores is the percentage of students who earned scores between 50% and 70%.
The area corresponding to a z-score of 0 is 0.5000, and the area corresponding to a z-score of 2 is 0.9772.
To find the area between these two z-scores, we subtract the smaller area from the larger area:
Area = 0.9772 - 0.5000 = 0.4772
Therefore, approximately 47.72% of students earned a score between 50% and 70%.