the distribution of data using standard deviation
Hello,
How much data should lie outside +/- 1 standard deviation from the mean in the following example?
If assessing the number of calories consumed by women in a day and the mean was 1800 and the standard deviation was 200, how many women consumed more than 1600 calories (population size 1000)?
What is the cutoff for the 99% points?
If I can get a detailed answer of how this is done, it would be greatly appreciated.
Thank you very much!
Re: the distribution of data using standard deviation
Is the data approximately normally distributed?
Re: the distribution of data using standard deviation
I think so. That's all the information I was given.
Re: the distribution of data using standard deviation
You should know that about 68% of the data lies within 1 standard deviation from the mean of any normal distribution...
Re: the distribution of data using standard deviation
Would it be possible for someone to answer this question in detail please?
Re: the distribution of data using standard deviation
Quote:
Originally Posted by
cupsk8
Would it be possible for someone to answer this question in detail please?
What a lazy and ungrateful poster. You're welcome...
Re: the distribution of data using standard deviation
That's really rude. I'm not being ungrateful or lazy. I don't understand this question at all and I have a huge exam tomorrow and this is the only example like this - it's not even a statistics exam and I've been looking at how to do it for about an hour now. I asked for help kindly AND said please.
Re: the distribution of data using standard deviation
I don't even want the ANSWER. I just want to know HOW I would go about doing it.
Re: the distribution of data using standard deviation
And I gave you the first step, and you did not even attempt to work through the problem, just begging for someone to do your work for you. I find that really rude, as do most of the helpers on this site.
Re: the distribution of data using standard deviation
Okay, can we just back up for a second. I don't even know where to BEGIN. Let me try using 68%. So, everything above 1600 calories would be 68% of the population. So then 680 women consumed over 1600 calories?
And I did say it would be greatly appreciated... I haven't done stats in 8 years.
Re: the distribution of data using standard deviation
Thank you for showing some effort. You are on the right track, but your question asked how many lie outside of +/- 1 standard deviation from the mean.
Now I'm just wondering if you made a typo. Are you sure the standard deviation is not 2000 calories instead of 200? In the context of the question, that would lie 1 standard deviation from the mean.
Re: the distribution of data using standard deviation
No, I checked and it's not a typo. That's what it asks. So is the answer not 680 women then?
Re: the distribution of data using standard deviation
Sorry I misread the question. My original hunch was correct.
No, the answer is not 680 women. Like I said, you are asked how many women consumed more than 1600 calories. All you have done so far is say how many ate between 1600 and 2000 calories (i.e. within one standard deviation from the mean).
Think about the bell curve. All you have shown so far is that 68% lie in between one standard deviation, but you haven't counted what is in the uppermost tail. See this diagram...
http://upload.wikimedia.org/wikipedi...iagram.svg.png
If 68% of the data lies within one standard deviation, then 34% lies within the tails. Half of that is in the uppermost tail.
So what percentage of the data are we actually counting?
Re: the distribution of data using standard deviation
32% of data should lie outside +/-1 standard deviation,
840 women consumed more than 1600 calories,
and cut-off points are 1300 and 2300 calories.
Re: the distribution of data using standard deviation
Quote:
Originally Posted by
cupsk8
32% of data should lie outside +/-1 standard deviation,
840 women consumed more than 1600 calories,
and cut-off points are 1300 and 2300 calories.
Yes, I agree you are dealing with (68 + 16)% = 84% of the data, which is 840 women.
You are also very close with your cutoff points, but it's actually 99.7% of the data that lies within 3 standard deviations from the mean, not 99%. To be more accurate you will need to convert to the standard normal distribution.