Suppose that on a certain examination in advanced mathematics, students from University A achieve scores which are normally distributed with a mean of 625 and a variance of 100, and that students from University B achieve scores which are normally distributed with a mean of 600 and a variance of 150. If two students from university A and three students from university B take this examination, what is the probability that the average of the scores of the two students from university A will be greater than average of the scores of the three students from University B? Suggestion: Find the distribution of the difference between the two averages.
To find the probability that the average of the scores of the two students from University A will be greater than the average of the scores of the three students from University B, we first need to find the distribution of the difference between the two averages.
Let's define two random variables:
X1 = score of student 1 from University A
X2 = score of student 2 from University A
Y1 = score of student 1 from University B
Y2 = score of student 2 from University B
Y3 = score of student 3 from University B
The average of the scores for University A, Ā, is given by:
Ā = (X1 + X2) / 2
The average of the scores for University B, B̄, is given by:
B̄ = (Y1 + Y2 + Y3) / 3
Since the scores are normally distributed, the sum of independent normal random variables is also normally distributed. Moreover, the mean and variance of the sum of independent random variables are the sum of their individual means and variances, respectively.
So, the mean and variance of Ā can be calculated as follows:
Mean of Ā = (mean of X1 + mean of X2) / 2 = (625 + 625) / 2 = 625
Variance of Ā = (variance of X1 + variance of X2) / (2^2) = (100 + 100) / 4 = 50
Similarly, the mean and variance of B̄ can be calculated as follows:
Mean of B̄ = (mean of Y1 + mean of Y2 + mean of Y3) / 3 = (600 + 600 + 600) / 3 = 600
Variance of B̄ = (variance of Y1 + variance of Y2 + variance of Y3) / (3^2) = (150 + 150 + 150) / 9 = 50
Now, let's find the distribution of the difference between Ā and B̄, D = Ā - B̄.
Since Ā and B̄ are normally distributed with known means and variances, D will also be normally distributed with the following parameters:
Mean of D = mean of Ā - mean of B̄ = 625 - 600 = 25
Variance of D = variance of Ā + variance of B̄ = 50 + 50 = 100
To find the probability that D > 0 (i.e., the average score of University A is greater than the average score of University B), we need to calculate the area under the distribution curve of D to the right of zero.
Using standard normal distribution tables or statistical software, we can find the probability of D > 0. Let's assume it is P(D > 0).
Therefore, the final step is finding P(D > 0) using the calculated parameters.