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# The figure shows a normal distribution with mean m and stand

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The figure shows a normal distribution with mean m and stand [#permalink]  07 Mar 2018, 15:34
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71% (01:42) correct 28% (02:42) wrong based on 14 sessions
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#GREpracticequestion The figure shows a normal distribution.jpg [ 22.71 KiB | Viewed 895 times ]

The figure shows a normal distribution with mean m and standard deviation d, including approximate percents of the distribution in each of the six regions shown. For a population of 800,000 subway riders, the numbers of subway trips taken per rider last January are approximately normally distributed with a mean of 56 trips and a standard deviation of 13 trips. Approximately how many of the riders took between 30 and 43 trips last January?

A 60,000
B 110,000
C 160,000
D 210,000
E 270,000
[Reveal] Spoiler: OA

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Re: The figure shows a normal distribution with mean m and stand [#permalink]  07 Mar 2018, 19:50
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The mean on a normal curve will be exactly in the middle, so on this curve the mean, or m, is 56. One standard deviation, or d, is 13 trips, so if we move to the left to m - d, we will be at 43 trips, the upper boundary of what the question is asking about. Moving a further 13 trips to the left will get us to m - 2d, which is thus 30. Since they're asking how many took between 30 and 43 trips, we know 14% of the population did so.

So what is 14% of 800,000? There are a couple ways to quickly estimate this. One is to know that 14% is about the same as 1/7th, and 1/7th of 800,000 should be a little over 100,000, since 1/7th of 700,000 would be exactly 100,000. Answer choice B looks pretty close.

Another way to estimate would be to say that 14% of a million would be 140,000, but our answer choice should be a little less than that. Again answer choice B is the only choice anywhere near our estimate.
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Re: The figure shows a normal distribution with mean m and stand [#permalink]  11 Jan 2019, 14:44
As it is between -2 and -3 sd the result is 14% of 8000000=112000==>110000 Answer. D.
Re: The figure shows a normal distribution with mean m and stand   [#permalink] 11 Jan 2019, 14:44
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