In this problem. We're looking at an instructor who has a class of students and is grading papers from the information that the problem provides. We know that these stand simple size is 40 students and that the mean grading time for each paper is six minutes, and the standard deviation for the grading time is also six minutes. Because the sample size 40 is greater than 30 we know that the central limit theorem also applies, and that means that the total time spent creating the papers, let's call it Big T is an approximately normal distribution. And because of that, we can then calculate the mean and the standard deviation for the total time distribution, which will be, well, Call me a T, and if I t. This is based on the sample size mean and standard deviation, so the equations will be and times the sample mean and the square root of n times the sample standard deviation, which, when you plug in the numbers, comes out to be 240 minutes, along with approximately a standard deviation of 37.9 five. So for the first part of the problem, we are trying to determine the probability that in grading all these papers and starting at 6 50 um, he will be able to watch his show, which will be at 11 o'clock if we transferred this or Trans, convert this to numbers or convert this two minutes. We are essentially trying to find the probability that the total time spent grading t is going to be less than 250 minutes, which is the time between 6:50:11 p.m. Which is when he would be grading and before the TV show starts. And using the central limit theorem and Z a standard normal random variable. We can then approximate this probability to be the probability that Z is going to be less than T minus the total meat divided by the total standard deviation. And once we plug in the numbers, these numbers, um, we have the probability of Z being less than 250 minus 2 40 which is the mean, um, time spent creating total time spent grading and 30 divided by 37.9 five. This comes out to be zero 0.2 six cool, and if we refer to the Z score table that the probability that we get from there is zero point 60 to 6, or approximately 60 0.3%. This means that the probability that the lecture or the instructor is going to be able to finish grading before his show starts is approximately 60.3% for the second problem. We are looking at the probability that he will miss time or essentially miss part of his show, and from the second part tells us that the show starts at 11 10 PM rather than 11 p.m. And based on that and converted to numbers, we are trying to find the probability that T will be greater than 260 minutes, which is the time between 6 50 11 10 PM, which then approximates to the probability that Z a random, a standard normal random variable will be greater than T minus mut divided by by T and plugged in that we get probability that Z is less sorry. Greater than 260 minus 2 40 divided by 37 0.95 Since we're looking at greater than we are looking at the right side of the normal distribution, which means that with respect to disease score, we're going to have one minus the probability for that Z score which calculated out, which is this value is going to be zero point five two seven and looking to the Z score table and looking at 0.53 on the Z score table, we find that this equates to one minus 0.7019 which comes out to be 0.2 98 one, which means that there is approximately a 29.8% chance that the instructor will miss part of his show while grading all of his papers.