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

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?

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

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