 Hi Pamela, this is Don again on your quiz on section 5.4 and 5.5. You did well on the quiz overall. This first problem that you missed has to do with the mean height of women in the country. That tells me this is the population, we're talking about all the women in the country. And their mean height is 64.3 inches. We took a sample of 75 women and we want to know what is the probability that the mean height in the sample is greater than 65 inches. We're given sigma which is the standard deviation of the population, 2.59. You got it wrong and you came up with an answer of .3824 and I'll show you why you got the wrong answer here. Let's look at Excel. Okay, here's Excel and it's a spreadsheet again. I think I shared it with the class version of it and I've built just a little table over here. The blue of course is my input and I put in the mean, 64.3, the population standard deviation 2.59, the end value for our sample of 75 women and the X that we're interested in is 65. Now here's where I think you went wrong. This information, the mean and the standard deviation are given for the population of women in the country. But we are interested in the sample of 75. So we can't use the sigma of 2.59. We've got to come up with the standard error or the population standard deviation. And we do that by using the formula standard deviation of the sample is equal to sigma divided by the square root of n. And that gives us a standard deviation of .299067 and I usually keep those many decimal places because it can make a difference. And I calculate my Z value using the standardized function of B4 which is, whoops, I've moved things down so I'll have to redo that. I use the standardized function which uses the mean, I mean the X value, the mean and the, let me just stop there. Okay, I apologize. I had at the last minute inserted these four rows so I could put the actual problem statement in here so I could remember it and reuse this later on. And that threw off these written equations out here so I just updated those so they're accurate. I apologize. But anyway, we use the standardized function and that takes in the X value, the mean of the sample, which remember under the central distribution theory, the mean of the sample for a normally distributed population is equal to the mean of the population. We need our standard error or standard deviation of the sample which we calculated again by taking the sigma and dividing by the square root of the sample size. Then I use the norm.s function which again takes the X value, the mean, the standard deviation and we use true because we want the cumulative distribution function which is all of the area under the curve to the left of our value of X and we want though the value to the right of X which is one minus that value, that's the area to the right and that is an answer of .0096. You use as I said the standard deviation of the population B6 and that gave a value of .6035, you subtract that from 1 to get .395. And again, I think you used slightly different technology and that is probably why there's a difference there in your value but I think that's what you did. Pamela, this is another problem that you had trouble with on that quiz and it has to do with use of the normal approximation to the binomial distribution and I am not really sure what you did wrong but let's go through and see how to solve this. We are given that 5% of the workers in the city use public transportation, we randomly select 265 and we ask them do they or do they not use public transportation so that fulfills the requirements for a binomial problem, binomial distribution problem in that we have randomly selected our workers that are independent and we are asking them a yes no question and the probability doesn't change for each of those. So let me bring up Excel, I have a spreadsheet that I built and I think I shared some of this with the class or it may have been another class if not I'll share the class after this and what I did there was bring over the given information we were given in a 265 and P of .05 we have to check N times P and N times Q to make sure they are both greater than 5 calculated Q which is just 1 minus P and both NP and NQ are much greater than 5 so we can use the normal approximation to the binomial. Some problems you may be asked for the mean of the normal approximation and the standard deviation and the formulas for those are just N times P for the mean and the square root of NPQ for the standard deviation of that normal approximation. So the first question was what is the probability of exactly 8 of the workers in our sample of 265 saying yes and when we use the normal approximation we have to apply the continuity correction which is adding and subtracting .5 and so our x1 value is 19.5 our x2 value is 20.5 and the way I label it x1 and x2 and I calculated the z value for each of those x's and got a z of 1.76 for the lower and 2.04 for the upper and then I used the norm s distribution function in Excel with a final parameter of true which gives me the cumulative area under the curve from left infinity up to that value of x or value of z I'm sorry and for the lower that's .96 for the upper that's .97 we want the area between because we want exactly 20 and it's that 19.5 to 20.5 band and I think actually in my stat lab that was the correct sketch and you actually selected the correct sketch subtracting those two gives us the area between x1 and x2 and that gives using Excel .0186 my stat lab answer was .0185 which is close to what I got with Excel and of course would be within rounding I would get ready for that you came up with .9714 and I don't know how you got that I I checked stat crunch just to see if let me go back here and put in 20 again and I checked a number of different possibilities actually all of them the greater than the less than and that was the close I've been wondering if you actually use the binomial distribution instead of the normal approximation to get your answer because that's awful close to .973 let's go on back to Excel the next question was what is the probability of at least eight workers said they used the public transportation using that same basic formula I got a z value and then this case because we've got an at least which is greater than or equal everything to the right of that value of z I use one minus the norm dis to get .9475 and that compares favorably with the .974 in MSL you got .0525 and I really don't know where that that came from if I put eight in the binomial and less than eight gives .083 that's not very close to .05 I put exactly eight I get .04 that's a little bit closer I don't know you know how you came up with that unless you were using some sort of the actual binomial instead of the normal approximation okay the last question was a part C I should say is if you what's the probability of less than 20 workers so I have a little setup there less than 20 we apply the continuity correction gives us down to 19.5 I calculate the Z for that and I put in use the norm s function again to calculate that area to the left of the Z .9609 both my stat lab and your answer were .9609 so you got that one correct the last part of the problem problem D says the authority offers discounts to companies have at least 30 employees to use public transportation your company this company has 468 employees was their probability that the company will not get the discount not as important there so if they are not getting a discount that means they have fewer than 30 because the the requirement is at least which is greater than or equal to so the complement of that which would not get the discount would be fewer than 30 or less than 30 so let's go back to our Excel and we've got to go back to this start again put in in now a 468 still a .05 would be our probability and we check in P and in two we're okay so I go down here to my less than block I put in 30 which means we apply the continuity correction down to an X2 of 29.5 that gives me a Z of 1.2932 which is on the upper side of the normal curve and using the norm s dysfunction I get a value of .902 90% chance that they will get fewer than 30 and that is the answer that my stat lab has you had .4661 I looked at step crunch the binomial and tried a number of different things to try to figure out how you got .4661 so I don't think you were using the binomial on this one unless you made a mistake but the answer is .9021 so I hope this helps a bit