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standard deviation..

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standard deviation..

by ska7945 » Fri Aug 22, 2008 10:50 pm
A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 68 percent of the distribution lies within one standard deviation d of the mean, what percent of the distribution is less than m + d ?

A. 16%
B. 32%
C. 48%
D. 84%
E. 92%


oa D
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Re: standard deviation..

by sudhir3127 » Fri Aug 22, 2008 11:05 pm
ska7945 wrote:A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 68 percent of the distribution lies within one standard deviation d of the mean, what percent of the distribution is less than m + d ?

A. 16%
B. 32%
C. 48%
D. 84%
E. 92%


oa D
its D 84

Since it is symmetric about m, it means, data less than m is 50% and data greater than m is also 50%.

thus each of the area from m to m+d and to m-d would be 34% (68/2=34%.)

m+d to m = 34%
data less than m is 50%
thus m+ d = 50 + 34 = 84%

hope that helps
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Re: standard deviation..

by fruti_yum » Thu Jul 30, 2009 2:44 pm
sudhir3127 wrote:
ska7945 wrote:A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 68 percent of the distribution lies within one standard deviation d of the mean, what percent of the distribution is less than m + d ?

A. 16%
B. 32%
C. 48%
D. 84%
E. 92%


oa D
its D 84

Since it is symmetric about m, it means, data less than m is 50% and data greater than m is also 50%.

thus each of the area from m to m+d and to m-d would be 34% (68/2=34%.)

m+d to m = 34%
data less than m is 50%
thus m+ d = 50 + 34 = 84%

hope that helps

I dont understand!!

Can someone explain how to tackle such problems?
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68-95-99 rule

by shahdevine » Thu Jul 30, 2009 3:03 pm
This is one of those things you will just have to tuck into back of your brain.

For every normal distribution, all values lie within 3 standard deviation of mean. Problem sets it up. Says population lies within one standard deviation. So first standard deviation would be m+d and m-d, which is (68%). Second standard deviation would be m+2d and m-2d, which is (95%). Finally, third standard deviation would be m+3d and m-3d, which is (99%). So since problem wants percent less than m+d it would be 68 percent plus 13% + 2% which equals approximately 84% for answer d.

you will have to google normal distribution curve to get visual on what i'm talking about. once you see it, it will all make sense.

cheers,

sd
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by tohellandback » Thu Jul 30, 2009 10:34 pm
distribution is symmetric about the mean m.
68 percent of the distribution lies within one standard deviation d of the mean

so 32 percent of the distribution lie outside that region.

check the image

add the numbers lessa than M+d and u have the answer
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