1. A study involving stress is done on a college campus among the students. The stress scores are known to follow a uniform distribution with the lowest stress score equal to 1 and the highest equal to 5. A sample of 75 students is randomly chosen. Let X = the stress score of one student. Let = the average stress score of 75 students. X ~ U(1, 5) mean = 3stdev = 1.1547

2. Find the probability that the average stress score for the 75 students is less than 2. Probability Statement: P(X-bar < 2) = 0.0000 Calculator steps: 2 nd,DISTR,normalcdf(-1E99,2,3,1.1547/sqrt(75)),Enter

3. Find the 90 th percentile for the AVERAGE stress score for the 75 students. 90% of the average stress scores are less than this value. Let k = the 90 th %ile Probability Statement: P(X-bar < k) = 0.90 k = 3.17 Calculator steps to find k: 2 nd,DISTR,invNorm(0.9,3,1.1547/sqrt(75)),Enter