Suppose that a New Jersey testing service has created a standardized test for assessing the reading and computing skills of 8th graders. The test consists of a reading part, which is structured to have a mean of 200 and a standard deviation of 50. It also has a mathematics part, which is structured to have a mean of 100 and a standard deviation of 20. Students take both parts of the test and are given a final score that is the sum of both parts together. What would be the standard deviation of the final student scores?
The mean combined score is 200 + 100 = 300 and the standard deviation of the combined score is sqrt[(50)^2 + (20)^2] = 53.9
My previous answer assumed the two score distributions are uncorrelated. In actual practice, this may not be true, becasus students who score low in one test would tend to be the ones who score lowest in the other, and similarly for the highest. In other words, the two distributions are correlated. The actual standard deviation is probably somewhere between the sum of the separate standard deviations (70) and the "root sum of squares (RSS)" number of 53.9 previously mentioned.
To find the standard deviation of the final student scores, we need to use some basic principles of statistics. In this case, we have two separate parts of the test, so we want to find the standard deviation of the sum of these two parts.
Let's assume X represents the reading score and Y represents the mathematics score. We know that the mean (μ) of the reading score is 200, and the standard deviation (σ) is 50. Similarly, the mean of the mathematics score is 100, and the standard deviation is 20.
To find the standard deviation of the sum of the two parts, we need to use the concept of variance. The variance (Var) of a sum of random variables is equal to the sum of their variances. Since standard deviation is the square root of variance, we need to square the standard deviations before summing them.
Mathematically, the formula for the variance of the sum of two random variables X and Y is:
Var(X + Y) = Var(X) + Var(Y)
So, let's calculate it step by step:
Step 1: Square the standard deviations:
Var(X) = σX^2 = 50^2 = 2500
Var(Y) = σY^2 = 20^2 = 400
Step 2: Add the variances:
Var(X + Y) = Var(X) + Var(Y) = 2500 + 400 = 2900
Step 3: Take the square root to find the standard deviation:
σ(X + Y) = √(Var(X + Y)) = √(2900) ≈ 53.85 (rounded to two decimal places)
Therefore, the standard deviation of the final student scores is approximately 53.85.