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# The height of women in Dewaria follows a normal distrubution

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GMAT Instructor
Joined: 04 Jul 2006
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The height of women in Dewaria follows a normal distrubution [#permalink]

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23 Sep 2006, 07:29
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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 6.3*10^-3% 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
Senior Manager
Joined: 11 May 2006
Posts: 253

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23 Sep 2006, 09:33
C

since height of 5 women is more than 4 SD of mean,

6.3/1000 % of population = 5
so population = 80000 approximately.
Manager
Joined: 29 Jul 2006
Posts: 83

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24 Sep 2006, 13:00
E

There should be 10 women, outside 4 times the SD, 5 above 184 cm and 5 below 136 cm.
Current Student
Joined: 29 Jan 2005
Posts: 5171

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24 Sep 2006, 15:53
kevincan wrote:
nischalb wrote:
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?
VP
Joined: 28 Mar 2006
Posts: 1351

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21 Oct 2006, 08:38
kevincan wrote:
nischalb wrote:
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
Senior Manager
Joined: 23 May 2005
Posts: 263
Location: Sing/ HK

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22 Oct 2006, 06:15
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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Manager
Joined: 04 May 2006
Posts: 169
Location: paris

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22 Oct 2006, 07:59
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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Senior Manager
Joined: 07 Jan 2008
Posts: 286
The height of women in Dewaria follows a normal distrubution [#permalink]

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20 May 2008, 11:10
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
Director
Joined: 23 Sep 2007
Posts: 767

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20 May 2008, 11:29
mbawaters wrote:
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".
Current Student
Joined: 11 May 2008
Posts: 552
The height of women in Dewaria follows a normal distrubution [#permalink]

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30 Jul 2008, 11:08
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?
Director
Joined: 27 May 2008
Posts: 539

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30 Jul 2008, 11:38
arjtryarjtry wrote:
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
Manager
Joined: 15 Jul 2008
Posts: 205

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30 Jul 2008, 11:58
durgesh79 wrote:
arjtryarjtry wrote:
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 ?
GMAT Tutor
Joined: 24 Jun 2008
Posts: 1346

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30 Jul 2008, 12:27
bhushangiri wrote:
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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Director
Joined: 27 May 2008
Posts: 539

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31 Jul 2008, 00:14
IanStewart wrote:
bhushangiri wrote:
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 )
Re: normal distribution   [#permalink] 31 Jul 2008, 00:14
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