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Re: The figure above shows a normal distribution with mean m an [#permalink]
Carcass wrote:
Attachment:
GRE The figure above shows a normal distribution with mean m and standard deviation d.png


The figure above 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 4000 students in a university, the heights of these students are approximately normally distributed with a mean of 67 inches and a standard deviation of 4 inches. How many of the students had heights between 71 and 75 inches?


A. 80

B. 560

C. 1120

D. 1360

E. 1920


We have \(m = 67\) and \(d = 4\).
Then we have \(71 = 67 + 4 = m + d\) and \(75 = 67 + 2 \cdot 4 = m + 2 \cdot d\).
Thus the proportion between \(71\) and \(75\) is the proportion between \(m + d\) and \(m + 2d\), which is \(14 \%\).
\(14 \%\) of \(4000\) is \(560\).

Therefore, B is the right answer.
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The figure above shows a normal distribution with mean m an [#permalink]
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The mean height is given as 67 inches and the standard deviation d is given as 4 inches. The interval between 71 and 75 inches is the interval between one and two standard deviations to the right of the mean. Since it is a normal distribution, 14% of the heights fall in this interval. That is 14% of 4000 students will have heights between 71 and 75 inches. That would be (14/100)* 4000 = 560 students. The answer is Choice B.
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The figure above shows a normal distribution with mean m an [#permalink]
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