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Re: On a particular test whose scores are distributed normally, [#permalink]
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sandy wrote:
On a particular test whose scores are distributed normally, the 2nd percentile is 1,720, while the 84th percentile is 1,990. What score, rounded to the nearest 10, most closely corresponds to the 16th percentile?

(A) 1,750
(B) 1,770
(C) 1,790
(D) 1,810
(E) 1,830


Explanation:
The diagram below shows the standard distribution curve for any normally distributed variable. The percent figures correspond roughly to the standard percentiles both 1 and 2 standard deviations (SD) away from the mean: The 2nd percentile is 1,720, roughly corresponding to 2 standard deviations below the mean.

Therefore, the mean –2 standard deviations = 1,720.

Likewise, the 84th percentile is 1,990: 84% of a normally distributed set of data falls below the mean + 1 standard deviation, so the mean + 1 standard deviation = 1,990.

Call the mean M and the standard deviation S. Solve for these variables:
M – 2S = 1,720
M + S = 1,990

Subtract the first equation from the second equation:
3S = 270
S = 90

The question asks for the 16th percentile, which is the mean – 1 standard deviation or M – S. (It’s a fact to memorize that approximately 2% of normally distributed data falls below M – 2S, and approximately 14% of normally distributed data falls between M – 2S and M – S.)

Since M – 2S = 1,720, add another S to get M – S:
(M – 2S) + S = 1,720 + 90 = 1,810

Notice that the percentiles are not linearly spaced. The normal distribution is hump-shaped, so percentiles are bunched up around the hump and spread out farther away.

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Re: On a particular test whose scores are distributed normally, [#permalink]
1
1 standaed deivation below the mean, so it is 16th percentile

so, If 1 SD below the mean leads to 16th percentile
so, SD for 15th ppercentile < 1 SD => 15/16 = 0.94
Now,
M - 0.94 X SD = 1900 - 0.94 X 90 = 1815.40, it is close to D
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Re: On a particular test whose scores are distributed normally, [#permalink]
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Re: On a particular test whose scores are distributed normally, [#permalink]
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