A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 68% of the distribution lies within one standard deviation d of the mean, what % of the distribution is less than m+d?
answer: 84%
help~~
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https://www.echidne-of-the-snakes.com/stats/stnorm3.png
This is based on the above graph for normal distirbution
The blue region is within one SD of mean ..
The light brown (pardon my colour sense) is within 2 SD ..and so on
On either side the blue region denoted m + d ..so the region less that m + d will be the rest 100 - 13.6 + 2.1 = 84 approx
This is based on the above graph for normal distirbution
The blue region is within one SD of mean ..
The light brown (pardon my colour sense) is within 2 SD ..and so on
On either side the blue region denoted m + d ..so the region less that m + d will be the rest 100 - 13.6 + 2.1 = 84 approx
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Perhaps a stupid question... if such a question is asked I assume that a graph must be provided as part of the question stem.sujaysolanki wrote:https://www.echidne-of-the-snakes.com/stats/stnorm3.png
This is based on the above graph for normal distirbution
The blue region is within one SD of mean ..
The light brown (pardon my colour sense) is within 2 SD ..and so on
On either side the blue region denoted m + d ..so the region less that m + d will be the rest 100 - 13.6 + 2.1 = 84 approx
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Well the graph is not provided ..if thats what u wanted to know ..and theres nothing too it ..just remember the graph ..and some mental calculation ....although ..it wud be better if GMAC doesnt give such stuff ..but anyways ..
I will try to figure out a diff way to explain
I will try to figure out a diff way to explain
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Well I think a graph would be really good here but it is not necessary to solve the question
coz the Q says "large population has a distribution that is symmetric about the mean m."
so 68% is within 1 sd of the mean i.e m+d, m-d =68%
remainig is 32% since the distribution is symmetrical abt the mean hence this 32% would be evenly divided on both the sides of the mean i.e 16% to the right & 16 % to the left
now the 16% to the right is greater than the mean & is also greater than m+d, so this is the portion which is only greater than m+d
hence the reqd % = 100 -16 = 84%
hope this is helpful
coz the Q says "large population has a distribution that is symmetric about the mean m."
so 68% is within 1 sd of the mean i.e m+d, m-d =68%
remainig is 32% since the distribution is symmetrical abt the mean hence this 32% would be evenly divided on both the sides of the mean i.e 16% to the right & 16 % to the left
now the 16% to the right is greater than the mean & is also greater than m+d, so this is the portion which is only greater than m+d
hence the reqd % = 100 -16 = 84%
hope this is helpful
Regards
Samir
Samir
Replying to an old thread. The below blog very nicely explains Standard Deviation concept:
https://amarsagoo.blocked/2007/09/m ... ation.html
https://amarsagoo.blocked/2007/09/m ... ation.html