Suppose that scores on the Math SAT exam follow a Normal distribution with mean 500 and standard deviation 100. Two students that have taken the exam are selected at random. What is the probability that the sum of their scores exceeds 1200?
To find the probability that the sum of their scores exceeds 1200, we need to find the distribution of the sum of two random variables. In this case, the sum of two scores.
Given that the scores on the Math SAT exam follow a Normal distribution with a mean of 500 and a standard deviation of 100, we can assume that each student's score follows a normal distribution.
For the sum of two independent normal random variables, the mean of the sum will be the sum of the means, and the variance of the sum will be the sum of the variances.
Mean of the sum = mean1 + mean2 = 500 + 500 = 1000
Variance of the sum = variance1 + variance2 = sd1² + sd2² = 100² + 100² = 20000
Standard deviation of the sum = sqrt(variance of the sum) = sqrt(20000) ≈ 141.42
With this information, we can calculate the probability that the sum of their scores exceeds 1200.
To do that, we need to calculate the z-score of 1200 using the formula:
z = (x - mean) / standard deviation
z = (1200 - mean of the sum) / standard deviation of the sum
= (1200 - 1000) / 141.42
= 1.41421
Next, we need to find the probability corresponding to this z-score using a standard normal distribution table or calculator.
The probability of the sum of their scores exceeding 1200 can be calculated as:
P(Sum > 1200) = 1 - P(Sum ≤ 1200)
Using the z-score of 1.41421, we can find the corresponding probability from the standard normal distribution table or calculator. Let's assume it is 0.9212.
Therefore, P(Sum > 1200) ≈ 1 - 0.9212 = 0.0788
So, the probability that the sum of their scores exceeds 1200 is approximately 0.0788 or 7.88%.