A distribution of test scores of 600 examinees follows a normal distriburion with an everage of 80 having a standard deviation 12, how many examinees would you expect to find above a score of 100?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of the Z score. Multiply by 600.
To find out how many examinees would be expected to score above 100, we need to understand the normal distribution and use z-scores.
Step 1: Calculate the z-score.
The z-score measures the number of standard deviations a particular data point is from the mean. We can calculate the z-score using the formula:
z = (x - μ) / σ
Where:
- x represents the value we want to find the z-score for (in this case, 100),
- μ is the mean of the distribution (80), and
- σ is the standard deviation (12).
Using these values, we calculate the z-score as follows:
z = (100 - 80) / 12
z = 20 / 12
z = 1.67 (rounded to two decimal places)
Step 2: Look up the z-score.
By referring to a z-score table or using statistical software, we can find the proportion of scores above a specific z-score. For z = 1.67, the proportion of scores above this z-score is approximately 0.0475.
Step 3: Convert the proportion to the number of examinees.
Since there are 600 examinees in total, we can calculate the number of examinees expected to score above 100 by multiplying the proportion by the total number of examinees:
Number of examinees = 0.0475 * 600
Number of examinees ≈ 28.5
Rounded to the nearest whole number, we would expect to find approximately 29 examinees scoring above 100.