The grades on the last math exam had a mean of 88%. Assume the population of grades on math exams is known to be normally distributed with a standard deviation of 6. What percent of students earn a score between 72% and 90%? (5 points)
0.6255
0.5843
0.6754
0.5346
To find the percent of students who earned a score between 72% and 90%, we need to calculate the z-scores associated with each of these scores and then find the area between these two z-scores.
Step 1: Calculate the z-score for 72%
The formula for the z-score is given by:
z = (x - μ) / σ
where x is the raw score, μ is the mean, and σ is the standard deviation.
For 72%, the calculation is:
z_72 = (72 - 88) / 6 = -2.6667
Step 2: Calculate the z-score for 90%
For 90%, the calculation is:
z_90 = (90 - 88) / 6 = 0.3333
Step 3: Find the area between these two z-scores
Now, we need to find the area between -2.6667 and 0.3333 on the standard normal distribution curve. This can be done using a z-table or a statistical calculator.
Using a z-table, we can find the area to the left of -2.6667 as 0.0036 and the area to the left of 0.3333 as 0.6293. To find the area between the two z-scores, we subtract the smaller area from the larger area:
Area = 0.6293 - 0.0036 = 0.6257
So, the percent of students who earned a score between 72% and 90% is approximately 0.6257 or 62.57%.
Therefore, the closest option to the correct answer is 0.6255 (rounded to four decimal places).
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to the Z scores.