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A researcher investigating mosquito populations recorded a

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A researcher investigating mosquito populations recorded a [#permalink]

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New post 14 Sep 2013, 20:38
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  65% (hard)

Question Stats:

59% (01:31) correct 41% (01:48) wrong based on 190 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
Spoiler: OA

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

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New post 14 Sep 2013, 23:53
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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.

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

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New post 15 Sep 2013, 03:17
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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.

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

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New post 02 Nov 2014, 20:34
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!)
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A researcher investigating mosquito populations recorded a [#permalink]

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New post 02 Nov 2014, 21:40
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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

Hence, Answer is B
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Re: A researcher investigating mosquito populations recorded a [#permalink]

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New post 01 Mar 2018, 14:13
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Re: A researcher investigating mosquito populations recorded a   [#permalink] 01 Mar 2018, 14:13
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