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# A researcher investigating mosquito populations recorded a

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Joined: 01 Sep 2010
Posts: 3380
A researcher investigating mosquito populations recorded a  [#permalink]

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14 Sep 2013, 19:38
1
12
00:00

Difficulty:

65% (hard)

Question Stats:

58% (01:55) correct 42% (01:56) wrong based on 205 sessions

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A researcher investigating mosquito populations recorded a list of estimates of mosquito populations at each of $$n$$ diﬀerent marshland locations in a certain country. The distribution of these estimates was symmetric about the average (arithmetic mean), and 70% of these population estimates were within one standard deviation of the mean. Of these estimates, 75 were more than one standard deviation below the mean. Approximately how many of the estimates were within one standard deviation of the mean?

A) 175
B) 350
C) 375
D) 425
E) 500

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Posts: 67
Location: United States
GMAT 1: 590 Q41 V29
GMAT 2: 540 Q44 V20
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Re: A researcher investigating mosquito populations recorded a  [#permalink]

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14 Sep 2013, 22:53
1
1
carcass wrote:
A researcher investigating mosquito populations recorded a list of estimates of mosquito populations at each of $$n$$ diﬀerent marshland locations in a certain country. The distribution of these estimates was symmetric about the average (arithmetic mean), and 70% of these population estimates were within one standard deviation of the mean. Of these estimates, 75 were more than one standard deviation below the mean. Approximately how many of the estimates were within one standard deviation of the mean?

A) 175
B) 350
C) 375
D) 425
E) 500

If 70% of population is within 1 standard deviation mean then 30% are more than 1 standard deviation below the mean.

75 are more than 1 sd below the mean therefore these 75 constitute 15% (Taking 15% below the mean and 15% above the mean)

Therefore total population is 500.

Therefore 70% of 500 is 350. who are within 1 standard deviation from the mean.

Hence B
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Joined: 02 Sep 2009
Posts: 61283
Re: A researcher investigating mosquito populations recorded a  [#permalink]

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15 Sep 2013, 02:17
1
1
shameekv wrote:
carcass wrote:
A researcher investigating mosquito populations recorded a list of estimates of mosquito populations at each of $$n$$ diﬀerent marshland locations in a certain country. The distribution of these estimates was symmetric about the average (arithmetic mean), and 70% of these population estimates were within one standard deviation of the mean. Of these estimates, 75 were more than one standard deviation below the mean. Approximately how many of the estimates were within one standard deviation of the mean?

A) 175
B) 350
C) 375
D) 425
E) 500

If 70% of population is within 1 standard deviation mean then 30% are more than 1 standard deviation below the mean.

75 are more than 1 sd below the mean therefore these 75 constitute 15% (Taking 15% below the mean and 15% above the mean)

Therefore total population is 500.

Therefore 70% of 500 is 350. who are within 1 standard deviation from the mean.

Hence B

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Hope it helps.
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Re: A researcher investigating mosquito populations recorded a  [#permalink]

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02 Nov 2014, 19:34
So guys, i did an error that 30% rather than 15% of population is at the left hand side of the distribution. So I guess in a normal distribution it must symmetry at both sides (duh!)
Manager
Status: I am not a product of my circumstances. I am a product of my decisions
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Location: India
Concentration: Operations, General Management
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WE: Operations (Energy and Utilities)
A researcher investigating mosquito populations recorded a  [#permalink]

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02 Nov 2014, 20:40
1
1
carcass wrote:
A researcher investigating mosquito populations recorded a list of estimates of mosquito populations at each of $$n$$ diﬀerent marshland locations in a certain country. The distribution of these estimates was symmetric about the average (arithmetic mean), and 70% of these population estimates were within one standard deviation of the mean. Of these estimates, 75 were more than one standard deviation below the mean. Approximately how many of the estimates were within one standard deviation of the mean?

A) 175
B) 350
C) 375
D) 425
E) 500

Let M = Arithematic Mean
70% Population Estimate is within 1 Standard Deviation of Mean (1D)
Let this 70% = X
i.e. (M-1D) <= X <= (M+1D)

70% of estimate is within 1 SD of Mean, it implies that 30% of estimates lie in the region on the left side of (M-1D) and on the right side of (M+1D).

Since, the estimates are Symmetric around the mean,
15% of Estimate is on left side of (M-1D), and
15% of Estimate is on right side of (M+1D).

Given, 15% of Estimate on left side of (M-1D) = 75

Therefore, X can be determined by:
15/70 = 75/X
X = 350

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Re: A researcher investigating mosquito populations recorded a  [#permalink]

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27 Aug 2019, 17:32
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Re: A researcher investigating mosquito populations recorded a   [#permalink] 27 Aug 2019, 17:32
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