Question

I NEED SMART PEOPLE, WILL GIVE MEDAL!


Lee missed the lesson on normal distribution and needs to do his homework. Explain to Lee how to use the mean and standard deviation of a normal distribution to determine the top 6% of the population.

Answer 1

and
Carl conducted an experiment to determine if the there is a difference in mean body temperature between men and women. He found that the mean body temperature for men in the sample was 91.1 with a population standard deviation of 0.52 and the mean body temperature for women in the sample was 97.6 with a population standard deviation of 0.45.

Assuming the population of body temperatures for men and women is normally distributed, calculate the 98% confidence interval and the margin of error for the mean body temperature for both men and women. Using complete sentences, explain what these confidence intervals mean in the context of the problem.

Answer 2

im desperate, i get the basics but don't know how to solve this

Answer 3

@terenzreignz

Answer 4

Yeah... sorry guys, stat isn't my strong point :)

Answer 5

its pre calc :( alrighty then

Answer 6

are you familiar with the second one?

Answer 7

Nope, sorry. Never really grasped stat D:

Answer 8

so what is your main question and i can try to help im good at standard deviation and such

Answer 9

great :) Carl conducted an experiment to determine if the there is a difference in mean body temperature between men and women. He found that the mean body temperature for men in the sample was 91.1 with a population standard deviation of 0.52 and the mean body temperature for women in the sample was 97.6 with a population standard deviation of 0.45.

Assuming the population of body temperatures for men and women is normally distributed, calculate the 98% confidence interval and the margin of error for the mean body temperature for both men and women. Using complete sentences, explain what these confidence intervals mean in the context of the problem.

Answer 10

MATH MALE YES!!!

Answer 11

Lee missed the lesson on normal distribution and needs to do his homework. Explain to Lee how to use the mean and standard deviation of a normal distribution to determine the top 6% of the population.

What methods have you used before on this type of problem?
Do you have a TI-83 or -84 calculator, and, if so, do you know how to use any of the STATistics functions?
Do you know what z-scores are and how to use them?
Do you have a table normal probabilities that would enable you to find the area/probability to the left of a certain z-score?

Answer 12

I'll try to build upon what you already know.

Answer 13

yes, i know most of what you said

Answer 14

i just don't have that calculator

Answer 15

No its not statistics its just Pre calc

Answer 16

your great thanks!!! :) @mathmale

Answer 17

you're

Answer 18

@yayo365, thanks for the enthusiastic welcome!
As a statistics problem, this one is fairly easy and straightforward. But if you're approaching the solution of this problem from a pre-calc standpoint, that somewhat limits the methods you can use. To help me know where to start, please share with me anything you've learned (or at least been exposed to) about finding areas under standard normal probability curves.

Answer 19

I note that the problem statement doesn't include any numerals, just the words "mean" and "standard deviation." That makes this an awfully general question; it'd actually be easier to solve a problem involving actual mean and std. dev. values.

Another question for you: are you familiar with the Empirical Rule?

Answer 20

ok the chapter focused on probability, z scores, mean, normal distribution, population percentages, and so forth. we mainly worked with the formulas to find margin of error and finding the z score.

Answer 21

YES and empirical rule!

Answer 22

After choosing a mean of 80 and a standard deviation of 8, I was able to determine that the top 6% of students would obtain scores of 92.438 or above. This took less than 1 minute on a TI-84 calculator.

I really need your help in understanding where you're coming from in terms of understanding statistics, so that I can build on your existing knowledge.

Answer 23

OK, then! So you have some knowledge of the Empirical Rule.
That rule states that 68% of your data will lie within one std. dev. of the mean; 95% will lie within two s. d. of the mean, and 99.7% within 3 s. d. of the mean. Sound reasonably familiar?

Answer 24

yes exactly :) thats how i learned it.

Answer 25

so how do we find 6% without the calculator?

Answer 26

you can use empirical rule and …. what else

Answer 27

i forgot what the other thing was, was it an equation?

Answer 28

standardizing a raw score?

Answer 29

Cool!!

So: If 95% of your data lies within 2 s. d. of the mean, that means that 100%-95%, or 5%, of your data lies outside 2 s. d. of the mean.

Half of that, on the right side of the normal curve, would be 2.5% of your data.

We are interested in "the top 6%." A very rough estimate would be that that top 6% is a little less than 2 std. devs. from the mean. How much sense does this make to you? Can you think of a way to improve upon this answer?

Answer 30

Actually, if you know about "standardizing a raw score," you know more about statistics than you've let on.

Standardizing a raw score implies finding a z-score.

What does "standardizing a raw score" mean to you?

Answer 31

so 6% is in the 3rd deviation? ok by standardizing the z scare I'm referring to the equation that you use to find the z score.

Answer 32

I'm going to summarize what we know already:
Since 95% of your data will be within 2 std. dev. of the mean, the area to the right of 2 std. dev. and under the normal curve is half of 5%, or 2.5%.
since 68% of your data will be within 1 std. dev. of the mean, the area to the right of 1 std dev. will be half of 68%, or 34%. That 6% mentioned in your math problem is closer to 2.5% than it is to 34%, right? So I'd conclude that the top 6% of your data would correspond to between 1.5 and 2 s. d. above the mean.

Answer 33

\[z=x-\mu/\sigma \]

Answer 34

ok, i understand that concept, but I'm just a bit confused on where you got the 5%

Answer 35

Building upon your knowledge of the standardized (or z-) score, \[z=x-\mu/\sigma \rightarrow z=\frac{ x-\mu }{ \sigma }, \] we look for the variables we know and the variables we don't know. We supposedly know the mean and the standard deviation. If we could find the z-score corresponding to the top 6% of scores, it'd be easy to solve your equation for x.

Let's stop a moment and see where we are. Does this discussion give you any idea regarding what to do next?

Answer 36

Empirical Rule states that 95% of your normally distributed data lies within 2 std. devs. of the mean, right?

Answer 37

yes of coarse

Answer 38

what is 100%-95%?

Answer 39

what i would think to do next is plug in my values, and solve the z score. then look on my new diagram with a mean of 80 and try to find the 6% pop

Answer 40

OHHHH ok haha

Answer 41

so thats what the 3rd deviation consists of

Answer 42

so our answer is in the 2nd deviation?

Answer 43

above 5% :)

Answer 44

Remember, @yayo365, that 5% is the data OUTSIDE of 2 std devs from the mean, that is, the top 2.5% and the very bottom 2.5%.

Your math problem mentions the top 6%, not top 2.5%, right?

i don't mean to go on and on; I know you want a conclusion.

What I'm trying to say is that the top 6% of your data will be within 2 s. d. of the mean, not within 3 s. d. I did a quick sample calculation on my calculator and found that the normalized score would be about 1.6.

Answer 45

so, do the best you can to tell me what you think this probelm statement is asking for and what remains to be found before you consider yourself finished with this problem.

Answer 46

For whatever it's worth, the standardized score corresponding to the TOP 6% of your data is z=1.555.

Answer 47

And that is independent of your mean and standard deviation.

So, you could conclude: "The top 6% of the population corresponds to a z-score (or normalized score) of approx. 1.555, and is represented by the area under the standard normal curve to the right of z=1.555."

Answer 48

Really, seriously, that's all you need to do, or can do. But somehow you have to be able to find that normalized score 1.555 yourself without a calculator or table; if you can't, then all you can say is that "the top 6% of the population lie to the right of a bit fewer than 2 standard deviations."

Answer 49

I know haha dw id rather learn it than just get an answer! yes that what I've caught so far, the answer is in the 2nd sd.

Answer 50

And obviously 1.555 s. d. is less than 2 s. d. above the mean.

Answer 51

z=1.5555 is the equation that we solved? AWESOME

Answer 52

I have this table that i use to find the z score

Answer 53

Let me reword your response a little: "the answer is in the 2nd sd." would be better if written, "the line separating the top 6% of the population from the lower 94% has a normalized score of a little less than 2."

Answer 54

(Shaking up yayo365 a bit): I had had no idea that you had this table!

Please look in the body of the table. find the value closest to 0.94 and specify what the corresponding normalized or z-score is. Hint: It'll be about z=1.6.

Answer 55

I've looked at the table myself and find that the nearest value in the body of the table to 0.94 is 0.9406. Can you agree with that?

Answer 56

whoops!! sorry about that. umm it says .82639

Answer 57

i go down on the table to .9 and then across to .04 to get .82639

Answer 58

But if we're talking about the top 6% of the population, @yayo365, we'd need to look for 0.94, which comes from 100%-6% = 94%.

Thank you for explaining what you were doing.

Your table of z-scores features z scores on the left margin and on the top margin, but the 0.94 we're looking for is not in either margin, but in the body of the table.