For this problem, we’re told that s, a t scores are normally distributed with a mean of 1518 and with a standard deviation of 325. In part, a were asked to find the probability that, if 1 s a t score is randomly selected, it will be between 42880 point so, in other words, we’re finding the probability of x being between 42880 point. So our first step here is to find z scores for both of these values using the red equation. I have on the screen. So, let’s start with 1480 using the equation. Z is equal to x, which is 1480 minus the mean which is 1518 divided by the standard deviation, which is 325 point that will give us a z score of negative .12. Now, let’s find our z score for 1440 point again using the equation. Z is equal to 1440 minus the mean divided by the standard deviation, and that will give us a z score of negative .24. So now that we have our 2 z scores, let’s find both of those in our z table over here on the left side of the screen. So we’ll start with a negative 0.12. So we’ll go down, on the left hand, column to negative 0.1, which is here and then over on the right to .02, which is up here when we meet in the middle we get .4522. Now, let’s find negative 0.24 go down to negative 0.2 over to .04 and we see that we get .405 twont now we know that our c table gives us everything that is less than what we’re trying to find. So this is everything less than 1480 point. This is everything less than 1440 to find. The probability of x being between 82840 will have to subtract this lesser value from this greater 1. So we’ll take .4522 minus .4052, which will give us .0478. So there’s a 4 or excuse me, .70. There’S a 4.70 percent chance of a randomly selected score being between 42880 point now in part b, we’re told that if 16 scores are selected at random, find the probability that their mean will be between 42880 point. So now, in part b we’re dealing with the sample and that sample size n we’re told, is 16 poi since we’re dealing with a sample and since our population of s a t scores we were told, is normally distributed. We can use the central limit theorem to…