a normal distribution has a mean of 20 and a standard deviation of 10. Two scores are sampled randomly from the distribution and the second score is subtracted from the first. What is the probability that the difference score will be greater than 5?
To find the probability that the difference score will be greater than 5, we need to consider the distribution of the difference scores.
Given that we have two scores sampled randomly, let's call them X and Y. Since they are sampled from a normal distribution with a mean of 20 and a standard deviation of 10, we can write their distributions as X ~ N(20, 10^2) and Y ~ N(20, 10^2).
To find the distribution of the difference score (Z = X - Y), we need to consider the properties of normal distributions:
1. The mean of the difference score is the difference between the means of X and Y: μZ = μX - μY = 20 - 20 = 0.
2. The variance of the difference score is the sum of the variances of X and Y: Var(Z) = Var(X) + Var(Y) = 10^2 + 10^2 = 200.
3. The standard deviation of the difference score is the square root of the variance: σZ = √Var(Z) = √200 ≈ 14.142.
Now, we can find the probability that the difference score (Z) will be greater than 5 by standardizing the distribution of Z and calculating the appropriate z-score:
z = (difference score - mean) / standard deviation
= (5 - 0) / 14.142
≈ 0.354
To find the probability associated with this z-score (z > 0.354), we need to consult the standard normal distribution table or use statistical software.
Using either method, we find that the probability P(Z > 0.354) is approximately 0.3632, or 36.32%.
Therefore, the probability of the difference score being greater than 5 is approximately 36.32%.