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Tro ovathoad reach distancos adult fomalos amo normal; dletelhaded MANAM 702 = Cu and candar davatlon 48 3 Cm Find the probabllily tnat an indlvidual dlstance greal...

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Tro ovathoad reach distancos adult fomalos amo normal; dletelhaded MANAM 702 = Cu and candar davatlon 48 3 Cm Find the probabllily tnat an indlvidual dlstance grealer Ihan 215,0cm Find the probability that tno mcan for 20 rndcmly solected distancos groalcr than 201.00 c Why Cin Ihe normal dietribluon bo Vsod Part (b} ovon Lhauah Ihe safple size dces not exceed 307The probabilty E (Round (ouf decimal plazes needed )The probabllity ! (Rourd Ghananecdod )Croose tno corroc { answor bolonTre nrma dis

Tro ovathoad reach distancos adult fomalos amo normal; dletelhaded MANAM 702 = Cu and candar davatlon 48 3 Cm Find the probabllily tnat an indlvidual dlstance grealer Ihan 215,0cm Find the probability that tno mcan for 20 rndcmly solected distancos groalcr than 201.00 c Why Cin Ihe normal dietribluon bo Vsod Part (b} ovon Lhauah Ihe safple size dces not exceed 307 The probabilty E (Round (ouf decimal plazes needed ) The probabllity ! (Rourd Ghana necdod ) Croose tno corroc { answor bolon Tre nrma disirblton canbe Used berause the ong nal population has norma G tnbuion Thu nonal dislnbulin cun used because the fnile popu abon correclon factor smiall; Tho normal dinlnbution can bo Usod bocauso tha probabilily Iass Ihan 0,5



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Heights of adult men between 18 and 34 years of age are normally distributed with mean 69.1 inches and standard deviation 2.92 inches. One requirement for enlistment in the military is that men must stand between 60 and 80 inches tall. a. Find the probability that a randomly elected man meets the height requirement for military service. b. Twenty-three men independently contact a recruiter this week. Find the probability that all of them meet the height requirement.

So we're going to have to distributions in this problem. One is the distribution for individual nails and looking at their heights, and we're assuming the main is 68 inches, with a standard deviation of three inches. So Approximately here is 71 1 standard deviation away, and the next one is going to have groups of size nine And that sampling distribution and then group of size nine, and then you find the main. So our sampling distribution is going to have a smaller standard deviation, still is going to be centered At 68", but now it's going to have a standard deviation Of three divided by the square root of nine, which is one. So our first question asked, what's the likelihood that you sample one person and you get that one person to be between 67 And 69 tall. And we can see that way, that means we would need to convert this to a Z value, so I have 67 minus 68 divided by the standard deviation. Now it's a Z value And I have 69 -68 Over three. And so we can see in both cases that we get uh negative one third, can I get positive one third? So we're symmetrical and I'm going to utilize my normal CDF for this And second and distribution normal CDF, and I'm going to use the lower value is negative .3 Repeating, and the upper is .3 repeating. And leave the mean at zero and a standard deviation at one. And when I do that, I find out that that probability comes out to be about 26%. Part B says now we're looking at this distribution and what's the likelihood of having the mean be between 67 and 69. And when we convert that into Z values again we take the number minus the mean, divided by the standard deviation, which is that one? Yeah. And this month groups this is the 69 minus this, 68 Do I, too? By that standard deviation up here, which is one. And so I get my Z values to be negative 1-1. Yeah, and again, I know that's about 68% but I'm going to use my normal CDF button again, normal cbf use my lower as negative one. My upper is one. And leave the mean at zero and standard deviation at one And we get .68, 2 rounds to seven. Now, part C. Yes, this answer is much larger, we expect that. But why? Basically? Because this standard deviation of this second problem is much smaller because this is the standard deviation that I use For groups of size nine. So we expect for those same numerical values to have a much higher percentage of the X bars being between these two numbers than individual values. So I didn't write all that down. But hopefully you see that this standard deviation smaller, therefore We would expect more values to be in between 67 and 69.

So we have our original population And we know that it has a mean of 205.5, Where the standard deviation of 8.6. And in our first question if of one individual person and the distribution is normal. Very, very important that we know that that what is the likelihood we randomly pick a person and the reach is greater than greater than 195. And let's just plop 1 95 here, we know it's lower than and we're finding a probability that is bigger than 950.5, converting that to a Z value 195 -205.5 Divided by 8.6. And let's see how nicely this comes out. Sometimes they do, sometimes they don't, And you're dividing by 8.6, it can end up coming out to be not so great. So we get that that Z value is about negative 1.22. And looking that up in our table, grab my book. Mhm. And I'm actually going to look up the probability of being less than 1.22. And when I do that, I get point, I get the point 8888. Now, part B, part B, we're dealing with a sample size. Uh Let's see how big that was. This time. Our sample size, I believe was 25. So we have a sample size of 25. And our sampling distribution, we're going to answer part C. Actually first our sampling distribution is why can we end up using a normal approximate or a normal distribution here, It's because the original Is normal. So it doesn't matter that this sample size is smaller than 30. So we have the mean of X. Bar is still going to end up being this 205.5. And the standard deviation of the expires each coming from a sample size of 25 is going to be that 8.6 divided by the square root of 25, which is five. And we want to find in part B What is the likelihood that we take 25 women and we end up getting that. The mean reach is greater than 203 and 203. We're just going to say is like here, Once again, we can see that that probability is going to be bigger than 50%. We convert it to a Z value. We know that's appropriate because of the original population again, was normal. Yeah, this should be two or 5.5. And then at 8.6 divided by, whoops. Just want that to be a five. All right, so let's see what that comes out to be. You know, the c value is negative And that's negative 2.5 on the top negative 2.5 on the top divided by and we have 8.6 divided by five. And when we get that value that comes out to be about negative 1.45 and I'm going to actually find the area below positive 1.45. And that comes out to 3.9265. Mhm. Yeah. And those are our answers always grab a picture. You can avoid making mistakes if you grab just a little archaic sketch.

So we are using the Idea that 31.6% of people older than 25 have only a high school diploma. And we're taking a random sample of 100. And we know that if we take N times P and N times one minus P, both these are going to be greater than or equal to five. So this distribution is approximately normal, and the mean of that distribution will be N times P, which will be 31.6, and the standard deviation will be the square root of n times P, which is this 31.6 times one minus or the complement of this. And square rooted. And that gives us about. Let me look on my work here, four 649 for the standard deviation. So that's what we're going to use to calculate our Z values And we want to find what's the likelihood you randomly pick 100 people and you get exactly 32. Now we could use binomial calculations, but we want to use approximate normal calculations. So we need to use that continuity correction And we will think 32 and go a little happy unit below. So we're going to go to 31.5 And then we're going to go a half a unit higher and go up to 32.5. And we want to convert these both to Z value. So I'm going to write this one down and then the next ones I won't. So 31.5 minus the mean divided by the standard deviation and now that's the z value And then the 32.5 minus the mean, divided by the standard deviation, It's a nine right there. And we do that calculation, we get this z value to be negative .45 And we get this see value to be .84. And then I can look up this value on my table and find the area below and this value on my table and find the area below and find the difference between those two. And when I do I end up getting oh I apologize these values are not right. Let me just erase those This value. I'm looking at the next question. This was negative .02. And we raise both these. This was negative .02. I'm sorry I was doing the next part because I already did the work on this. And then look up this file, you find the area below, look up this value, find the area of the below. And when you subtract those two you get .0833. There we go. Now this is correct. Now part B. We're trying to find from 30 out of 100 all the way up to 35. And again we could find 30 31 32 33 34 35. And use our binomial calculations, but we want to use that continuity correction and so we need to bump this down, that's what we're going to use to calculate the Z. And we need to bump that up. Now, when I take this value right here and basically plop it right here in my calculator and to convert it to a Z value, I find that I get This z value is what I wrote down before, is the negative .45. Yeah. And again take this value and plop it in place of this value. Take to subtract the mean away and divided by the standard deviation. And we'll find out that this one comes out to be .8 for and then by the area below here, Find the area below here and subtract those two values and that gives us an area of .4732. And then our last one Is to find what's the likelihood of having more than or equal to greater than or equal to or at least 25 is the way I think it was worded. Now we need to bump that down to use the continuity correction. And so when I convert this to a Z value again, using that same calculation, this minus the 31.6, and then divided by the standard deviation of About 4.65. That gives me a Z value Of -1.53. And then instead of finding the area, instead of looking up negative 1.53 and finding this area, I actually looked up positive 1.53 and found the area below. So I didn't have to do this attraction. And that gives me a .9370. No, and I think we have all three parts.

So we have our overhead reach problem. Again, we know that our original population is normal and has that mean of 205.5 and a standard deviation of 8.6 and these are in cm and we want to know what's the likelihood that we pick an individual person and the reach is greater than 218.4. So again, that's up here Someplace and we're finding that area bob and converting that to a Z value, Taking the 218.4 -205.5 Over 8.6. And again we can do this because the original population is normal And we find out that that Z value comes out to be 1.5 And that corresponds with the probability of .066, 8, I believe. Now we have Our other question for part B and Part B has a sample size of nine and we know that because this original distribution is normal, we're going to answer answer actually questions see right now. So yes, we can use approximately the normal distribution because the original population original pop is normal. So the sampling distribution will be normal regardless of our sampling size. So The mean of expires will still be 205.5 and the standard deviation of expire will be that 8.6 divided by the square root of nine, which will be three. So we want to know what's the likelihood of getting a group of nine people having a mean That is greater than 204. And so we can convert that to a Z value 204 -100205.5 Divided by the 8.6 over three. And let's put that in the calculator to hold four -2055. But I'm kind of machine myself that I didn't just subtract that in my head, but I already typed it in and be careful about plugging it inappropriately. And we're going to get that that Z value is approximately negative 0.52. And so we can either use our table or use our normal CDF depending on what you're allowed to do and putting that into my calculator. I'm putting it into my calculator, I'm finding that, wow, Oops, and I need to do one minus that answer. I kind of put it in the wrong way, I end up getting about .698 five For that area. And again when we look at where this value is here, we know that that area is going to be bigger than .5. I always recommend you draw a little picture and sketch because you can catch mistakes um that you may use on your calculator or reading the table. So draw a picture


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