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10 Feb 2017, 01:20
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
43% (02:00) correct
57%
(02:08)
wrong
based on 56
sessions
History
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Result
Not Attempted Yet
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
A) 65
B) 48
C) 24
D) 15
E) 7.5
11 Feb 2017, 01:10
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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.
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.
Attachments
Normal-distribution-curve.jpg [ 26.46 KiB | Viewed 2296 times ]
11 Feb 2017, 02:54
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duahsolo
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
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.
