biology-the-heights-of-men-in-a-survey-are-distributed-normally-about-the-mean-use-the-graph-for-exe

Question

Biology The heights of men in a survey are distributed normally about the mean.
Use the graph for Exercises 1 $-4 .$
About what percent of men aged 25 to 34 are $69-71$ inches tall?

Natalie Anderson · Accepted Answer

okay. About what? Percent of men aged 25 to 34 are, uh, 69 December 1 inches tall. First thing you need to do is practice reading graphs, notice that 69 gets in this column 70 and 71. So you&#x27;re gonna have to figure out what these percentages represent. So this is 10 and 20 then this must be 12 14 16 18. So you&#x27;re going to have to add up all of these pieces. So, for example, 69 is that See this stuff? So probably 13%. Our 69 inches 70 inches is almost see 12 14 16%. And 71 is, um, probably 14% ish store best approximations. So we just need to do 13 16 plus 14 to get roughly 43%. The yeomen are between 69 71 inches tall.

Abigail Carden · Answer

So here, using our graph, we want to find the percentage of men that are less than 70 in just haul, which means we want to include heights from 64 inches all the way up to 50 69 inches. So we&#x27;re going to estimate the bars on our graph to find what percentage of men are at each height increments and then add them all up. So it&#x27;s 69 inches. That is equal to about 13% of men. 68 inches is about 11.5%. At 67 there are about 6% of men that bowline that height range. At 66 inches, it&#x27;s about 5%. 65 is about 3% and 64 is about 1%. So we want to add all of these percentages up to find the total percentage, which comes out to 39.5% which is about 40%

Natalie Anderson · Answer

if he did the previous problem. I think those two problems ago you&#x27;re supposed to find out what percent of men were between these inches. And you found that there were. 43% of the men are between 69 71 just home. So that told you if there were 100 men, 43 of them would be got tall?

Abigail Carden · Answer

you&#x27;re asked to find what percentage of men are between one standard deviation above and once they&#x27;re deviation below the mean. So we know that our mean is equal to 70 are standard deviation is equal to 2.5 and we&#x27;re going to use the formula. Z is equal to ah value minus the mean over standard deviation to find where ah one and negative one standard deviations would be from army. So to start will set one equal to Z because we want one center deviation above one is equal to you. Minus are mean of 70 over 2.5, and that gives us that you was equal to 72.5. So it&#x27;s once they&#x27;re deviation above our means. For once in your deviation below will set C equal to negative point, which gives us negative One is equal to B minus 70 over 2.5. Therefore, V is equal to 67.5, so 72.5 is above and 67.5 is one standard deviation below. So that means we&#x27;re looking at the following heights. We want to look at heights from 68 inches from 2 72 inches to find where these men lay and what percentage they encompass. So our percent is equal to 11.5% from our graph for 68 inches, plus 13% for 69 inches, plus 16% for 70 inches. Waas, 14% for 71 inches, plus 12% for 72 inches and adding that up, we get 66.5% which is close to the 68% that is expected within one standard deviation above or below army.

Gatjang Deng · Answer

Okay, So for this problem, we were told that, dear. According to the health survey, the Iight the height of adult males in the United States are normally distributed was mean of 69. It&#x27;s not inches and 28 inches. So so we&#x27;re given mean, so mean We don&#x27;t use a good meal because 69 inches and then you given the this standard deviation. It&#x27;s 28 sanitation where you would use sigma because 2.8 inches. So that&#x27;s how you define. So so the survey is normally be modeled. Relax. Is it because 21 over and in the square to pi times sigma time Miss E to the minus 1/2 X minus mu. It was Sigma squared. And this is true for all ex. This is true for all ex great, and you questions you so that means effort max zero. Every else in that domain you can plug in Miracle 69 and Sigma equals 2.8 inches to this expression here. Now the question would like. The answer is to determine the probability that an adult male shoot at random chosen a random will be between 65 inches and similar three inches tall. Hey, it&#x27;s similar to say What is P? Alexe is between, uh, 65 and 73 inches tall. So, mathematically, this is you. We can translate this into a little trouble. We will become so limits of integration are gonna be from 65. Consider it. It&#x27;s free after lex D X. So all right. Now you can plug in old in America about stuff after we see the ready. So mu is 69 signals to plate into f of X. Now, for our into go, um, interval will become one over to two pi and then square root every signal. What is Cigna Signature stamp? Civilian. We used to Quinn eight inches e to the minus 1/2 in the X minus 69 of a sigma, which is 2.8. And then you square the entire thing. The X So that&#x27;s the interval we need to evaluate. And now limits are from 65 two 73. Is the car into a goal we&#x27;re supposed to do? You should see that I pushed out this constant fear. One over to Red Troop I, Thomas 2.8. Because of that Constance, I pushes how it automatically So the only part of integral you have to do is this part here, and news board is no elemental way to do it. Um, you can use but war from which is freely available on the Web. Or you can use my people. The people are going to uh, So anyway, So if you put this into an integral calculator, uh oh, are integrating numerically for them. We&#x27;re using that Simpson&#x27;s rule. You will get something like 0.8469 And so or roughly roughly rather the 85% 85% now moving on to part B. So before B, we would like to know the percentage of the off the adult male population that is more than six feet six feet. Remember, You know, the units were using the interest, so we needed to hurt six feet. So how many in deciding six feet so six feet. He&#x27;s in close to 12 every six inches because one foot one foot it was 12 inches. So you might plan six by 12. You get 72 senator two inches. Um, so now that so the question becomes, what is the probability was a problem. X rated integrity, too. On translating this integral, it will be Seamus integrating from 72 to infinity of f of x d x. When we put in everything you should get inta go. And he&#x27;s going to be the constants that before So one of a square to pie Time of sigma, which was 2.8 and then the inter go from 72 to infinity. Then you have ye to the miners 1/2 x minus 6.69 point zero of a 2.8. Okay, then the whole thing squared dx again. This integral has no elemental form. This is how you do the interval. Now there are trucks of doing intervals like that spirit. I&#x27;m not gonna go into that. Um so using Wolfram maple or what about Integral Cog? Yeah. If you should get something like zero point 14 or 14%. 14% questions asking for an answer in percentage to. So you get 14%. And that&#x27;s how you have that question

Tony Wilson · Answer

So one of three wants you to come up with the sample size to be 93% confidence and within one inch. So we want to be within one inch 93%. Confidence gives us a critical value of 1.812 times the standard deviation over the square root of that. So if you square everything multiple, I&#x27;m beside my end. Divide by one, you will end up with and equals one point 812 square times 2.5 squared over one. Swear this gives you 20.52 You always around these up, so you would need 21 males.

Problem

Biology The heights of men in a survey are distri…

Problem 1 Easy Difficulty

Biology The heights of men in a survey are distributed normally about the mean.
Use the graph for Exercises 1 $-4 .$
About what percent of men aged 25 to 34 are $69-71$ inches tall?

Answer

43$\%$

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