In a psychology class of 100 students, test scores are normally distributed with a mean of 80 and a standard deviation of 5. Approximately what percentage of students have scores between 70 and 90?
A. 68%
B. 80%
C. 95%
D. 99%
To find the percentage of students with scores between 70 and 90, we need to calculate the z-scores for these values and then use the standard normal distribution table.
First, we need to convert the scores 70 and 90 into z-scores. The formula to calculate the z-score is:
z = (X - μ) / σ
Where:
X is the score we want to convert,
μ is the mean of the distribution, and
σ is the standard deviation of the distribution.
For X = 70:
z1 = (70 - 80) / 5 = -2
For X = 90:
z2 = (90 - 80) / 5 = 2
Next, we can use the standard normal distribution table (also known as the Z-table) to find the area between these two z-scores. The Z-table provides the percentage of the distribution that falls below a specific z-score.
Looking up the z-scores -2 and 2 in the Z-table, we find that the corresponding areas are approximately 0.0228 and 0.9772, respectively.
To find the percentage of students between scores 70 and 90, we subtract the area found for the lower z-score from the area found for the higher z-score:
Percentage = (0.9772 - 0.0228) * 100 = 0.9544 * 100 ≈ 95.44%
Therefore, approximately 95% of students have scores between 70 and 90.
The correct answer is C. 95%.
12
I Do not know
You haven't memorized the proportion within 2 standard deviations from the mean?
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 the Z score.