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kevincan
nischalb
E

There should be 10 women, outside 4 times the SD, 5 above 184 cm and 5 below 136 cm.

Good point! OA=E


Originally chose (C), but now (E) sounds plausible becuase we are working with a "normal distribution." The five women over 184cm are not "outliers" and therefore need to be balanced off with five women under 136cm to achieve the stated mean of 160.

10=.0063*0.01x

Why doesn't the straightforward math work out? Missing a decimal somewhere?
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kevincan
nischalb
E

There should be 10 women, outside 4 times the SD, 5 above 184 cm and 5 below 136 cm.

Good point! OA=E


can someone explain this please...Im lost
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How do you work out 10^-3%?

I know it equals 1/ 10^(3/100) but don't know what to do after that...
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Let 6.3*10^-3% =A%, so A%=6.3/10^3

so A%=10 people
since A%=A/100, so A =10^3*(10/6,3)*10^2=1,6*10^5=160 000

hope it helps...
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The height of women in Dewaria follows a normal distrubution with mean 160 cm and standard deviation of 6 cm. In a normal distribution, only 0.0063% of the population is not within 4 standard deviations of the mean. If 5 women are more than 184 cm tall, approximately how many women live in Dewaria?

(A)16,000
(B)40,000
(C)80,000
(D)100,000
(E)160,000
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mbawaters
The height of women in Dewaria follows a normal distrubution with mean 160 cm and standard deviation of 6 cm. In a normal distribution, only 0.0063% of the population is not within 4 standard deviations of the mean. If 5 women are more than 184 cm tall, approximately how many women live in Dewaria?

(A)16,000
(B)40,000
(C)80,000
(D)100,000
(E)160,000

E

5 wowen are taller than 4 standard deviations. But with a normal distribution, 5 women should also be shorter than 4 standard deviations. 10 women is 0.0063% of the population, approximately 160000 women in the population.

tricky question, I forgot about the other side of the distribution the first time I tried to solve the problem. This is not so much a math question, it is a "test your attention to details question".
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The height of women in Dewaria follows a normal distrubution with mean 160 cm and standard deviation of 6 cm. In a normal distribution, only 0.0063% of the population is not within 4 standard deviations of the mean. If 5 women are more than 184 cm tall, approximately how many women live in Dewaria?

(A)16,000
(B)40,000
(C)80,000
(D)100,000
(E)160,000

is not within one D, 68%, within 1D to 2D, 28% and 2D to 3D , 4% occur? so here , the given sample lies outside the normal curve is it?
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arjtryarjtry
The height of women in Dewaria follows a normal distrubution with mean 160 cm and standard deviation of 6 cm. In a normal distribution, only 0.0063% of the population is not within 4 standard deviations of the mean. If 5 women are more than 184 cm tall, approximately how many women live in Dewaria?

(A)16,000
(B)40,000
(C)80,000
(D)100,000
(E)160,000

is not within one D, 68%, within 1D to 2D, 28% and 2D to 3D , 4% occur? so here , the given sample lies outside the normal curve is it?

5 women more than 184 ( 160 + 4*6 : mean + 4*SD)
5 women must be less than 160-24 ... since normal distribution is same on both sides of mean

total women outside 4*SD = 10 as per Question stem this is 0.0063% of total population X
X * 0.0063 % = 10
X = 10000000/63 = 160,000 .... Option E
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arjtryarjtry
The height of women in Dewaria follows a normal distrubution with mean 160 cm and standard deviation of 6 cm. In a normal distribution, only 0.0063% of the population is not within 4 standard deviations of the mean. If 5 women are more than 184 cm tall, approximately how many women live in Dewaria?

(A)16,000
(B)40,000
(C)80,000
(D)100,000
(E)160,000

is not within one D, 68%, within 1D to 2D, 28% and 2D to 3D , 4% occur? so here , the given sample lies outside the normal curve is it?


5 women more than 184 ( 160 + 4*6 : mean + 4*SD)
5 women must be less than 160-24 ... since normal distribution is same on both sides of mean

total women outside 4*SD = 10 as per Question stem this is 0.0063% of total population X
X * 0.0063 % = 10
X = 10000000/63 = 160,000 .... Option E


While i agree the answer is 160K, I am not sure if gmat expects you to know the symmetrical nature of normal distribution. The question should have explicitly stated that normal dist. is symmetrical about the mean.
Don't you agree ?
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bhushangiri

While i agree the answer is 160K, I am not sure if gmat expects you to know the symmetrical nature of normal distribution. The question should have explicitly stated that normal dist. is symmetrical about the mean.
Don't you agree ?

Yes, I agree- you aren't expected to know the properties of the normal distribution on the GMAT. I'm curious where the original question is from.
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IanStewart
bhushangiri

While i agree the answer is 160K, I am not sure if gmat expects you to know the symmetrical nature of normal distribution. The question should have explicitly stated that normal dist. is symmetrical about the mean.
Don't you agree ?

Yes, I agree- you aren't expected to know the properties of the normal distribution on the GMAT. I'm curious where the original question is from.


most probably this is a question from India's MBA entrance exam for IIMs ... the place mentioned in the question "Dewaria" is an actual place in India .... (not well known though )



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