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Last year at a particular day care center, 98% of the children were si

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Last year at a particular day care center, 98% of the children were si [#permalink]

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New post 10 Feb 2017, 00:20
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Question Stats:

34% (01:30) correct 66% (01:51) wrong based on 29 sessions

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Last year at a particular day care center, 98% of the children were sick less than 40 days and 50% of the children were sick less than 25 days. If days sick was normally distributed among the children, what is the standard deviation?

A) 65
B) 48
C) 24
D) 15
E) 7.5
[Reveal] Spoiler: OA

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Re: Last year at a particular day care center, 98% of the children were si [#permalink]

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New post 11 Feb 2017, 00:10
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The pattern follows the normal distribution.

50% children are sick less than 25 days = i.e. 1 SD Approx (it should be 68% if it follows normal distribution);
98% children are sick less than 40 days = i.e. 2 SD Approx (it should be 95% if it follows normal distribution);


Therefore, the difference is 40 -15 = 15% = this includes 1 SD from both side of the median (refer to the graph);
Hence, 1SD = 15/2 = 7.5 %;
Answer E.
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Re: Last year at a particular day care center, 98% of the children were si [#permalink]

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New post 11 Feb 2017, 01:54
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duahsolo wrote:
Last year at a particular day care center, 98% of the children were sick less than 40 days and 50% of the children were sick less than 25 days. If days sick was normally distributed among the children, what is the standard deviation?

A) 65
B) 48
C) 24
D) 15
E) 7.5


Normal distribution and its 8.2-95.4-99.7 rule are NOT tested on the GMAT.
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Re: Last year at a particular day care center, 98% of the children were si   [#permalink] 11 Feb 2017, 01:54
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