Problem-14 A Six- ~Sigma Black Belt is studying the performance of financial analysts from a bank using their scores on a certain aptitude test She has collected d...

A population has mean 72 and standard deviation 6 . a. Find the mean and standard deviation of $x-$ for samples of size $45 .$ b. Find the probability that the mean of a sample of size 45 will differ from the population mean 72 by at least 2 units, that is, is either less than 70 or more than 74 .

For this problem, we're going to start by finding our cumulative relative frequency distribution. Um, we already found the relative frequency distribution in the previous question, So I already copied that information down, and we're just going to fill in the cumulative relative frequency column. So to get the cumulative relative frequency for the class 0 to 4, it's the same as the relative frequency, which is 40.8 for the Class four through eight. We add the previous cumulative relative frequency of 80.8 to the classes relative frequency 0.16 So that's going to give us 0.24 and we continue to add the previous cumulative relative frequency to the new classes relative frequency. So for eight through 12 we're going to do point to four plus 40.16 which is going to give us 0.40 for 12 through 16. We're gonna do 160.40 plus 0.40 which would give us 0.80 for 16 through 20. We do 200.80 plus 0.1 to which gives this 0.92 for 23 24. We do 240.92 plus point of six, which gives this 60.98 and last class 24 through 28 we add 280.98 and 0.2 which gives us 1.0. Now, we're gonna use this cumulative relative frequency distribution to construct our dog. I've and part B. Now we're going to construct are all guys. So we start by labeling the X and Y axis. The X axis is test scores and we use the class boundaries. And then the cumulative relative frequency goes on the Y axis. And of course, we've titled our graph at the top. So now, to create our archive for each class, remember that the cumulative relative frequency is plotted at the upper boundary of each class. So for the cause, 0 to 4, we start at 00 and then we plot the cumulative relative frequency of 0.0 a at the four. And then we connect those with a wine for the class four through eight. We plot the key months of relative frequency of 0.24 at the A because that's the upper class boundary, and we again connect with a line. So then, at 12 we plot the 120.40 connect with the line at 16 we plot 160.80 at 20. Re plot the 200.92 At 24 we plot the 240.98 and at 28 we get to 1.0. Where are five should end and that is the AWG I for our data. Now we're going to use our archive to answer part C. So for part C, we are going to look for what percent have no more than 16. So we go to 16 on our archive and we're gonna go up to the line from there, and then we're going to go over and see what the value is. And we know when we constructed the AWG, I've at 16 we plotted a cumulative relative frequency of 160.80 But one thing to keep in mind is that that 10.80 doesn't include the value 16. And when we talk about no more than 16 16 is included. So we're thinking with no more than 16 you're thinking less than or equal to 16. And so that percentage right there, 80% doesn't really include he 16 necessarily. But it's a good approximation, so we can say that approximately 80% of students scored no more than 16

In this question were given a mean and sample standard deviation for some test scores and asked to find the Z scores for several individual data values. So let's say we have the mean of 74.2 on a sample standard deviation of 11.5. If you recall the formula for finding a Z score is X minus X bar over s where X is the data value. X bar is the sample mean and s is the sample standard deviation. So for our first part of the question, we are given us test score of 54. So somebody who performed pretty well below average So to find there's you score will substitute all of our known information into the formula for the Z score and divide by our standard deviation. So subtraction first, following order of operations and then division gives us a Z score of negative 1.76 So somebody who scored between one and two standard deviations below the mean Our second test score is 68. So somebody who scored below the mean but not as far below the mean, so we would expect disease score below zero, but not as far below zero. So 68 minus 74.2 over 11.5. And we find that that Z score is in fact negative. But on Lee negative 0.54. What if somebody performs better than average? So what if somebody got a test score of 79 so a little bit above average. So we would again substitute all of our known information into the formula and divide by our standard deviation here. And we're going to expect a Z score above zero, but not a lot above zero, because this score is not very far above average. So that's a Z score of 0.42 and then finally parte de what if somebody did even better above average? What if they scored in 93? So we would expect again a positive Z score because the scores above average, but it's even further above average. So even higher number and this ups supposed to be a five, not a 11.2 11.5, and this sea score turns out to be 1.63

So in this question, we're told that scores in an exam are normally distributed with me 382 and standard deviation 26, And we're justifying the score that is in the 50th%ile. So that's basically this score right here, 58%ile, so that C0 point Hi, so probably that Z is less than the star is point five, So these are in our case is zero. And so that translates to a score x star of 382 45. Okay, for being with us to find the score that's in the 90th percentile, So that's followed easy, less than Z Star is equal to not 0.9. And for that our value of Z star is 1.28, which converts to an ext r value of 415 work part.