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On a particular test whose scores are distributed normally, the 2nd pe  [#permalink]

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Question Stats: 29% (02:29) correct 71% (02:28) wrong based on 50 sessions

### HideShow timer Statistics 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

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Posts: 3087
On a particular test whose scores are distributed normally, the 2nd pe  [#permalink]

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Gnpth 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

Attachment: normdist.png [ 13.13 KiB | Viewed 1759 times ]

Quickly sketch a normal distribution: 0 ----2----16----50----84----96----100

A score at 84th percentile is three standard deviations away from a score at 2nd percentile

1,990 - 1,720 = 270

$$\frac{270}{3 SDs}$$ = 90 = one SD

16th percentile score is +1 SD greater than 2nd percentile score
16th percentile score would be:
1,720 + 90 = 1810

Explanation
Data is clustered around the mean, and percent distributed in any segment has a fixed number

"Distributed normally" means the data fall into the normal distribution curve, the top figure in the diagram.
68 percent of the data fall within one standard deviation from the mean -- one deviation above, one deviation below (so 34%/34%)

The percent of data underneath any part of the curve always follows the distribution seen in the top figure (figures in green ink):
2 (percent of the data), then 14 (percent of data) , then 34, 34, 14, 2
Memorize these numbers 2, 14, 34 <-[mean] -> 34, 14, 2

If you memorize those numbers and start with 50 as the mean, just add to get percentiles:
RIGHT OF THE MEAN: (50+34) = 84 | (84+14) = 98 | (98+2) = 100
LEFT OF THE MEAN: (50-34) = 16 | (16-14) = 2 | (2-2) = 0

Standard deviation corresponds with percentiles in a normal distribution
See middle figure, that pattern will always hold:
0th percentile is -3 SDs from mean
2nd percentile is -2 SDs from mean
. . .
84th percentile is + 1 SD from mean
100th percentile is +3 SDs from mean

What is score at 16th percentile here?

Score at 2nd percentile = 1,720
Score at 84th percentils = 1990

Score at the 16th percentile?

The score at the 84th percentile is THREE standard deviations from the score at the 2nd percentile (just count segments)

Find the difference between the scores
Divide by 3 (for three standard deviations)

(1,990 - 1,720) = 270

$$\frac{270}{3SDs}$$ = 90

So 90 = ONE standard deviation

The 16th percentile is +1 SD away from the 2nd percentile (see top and middle figures)

So, 16th percentile score = the given 2nd percentile score (1,720) + ONE standard deviation (90)

1,720 + 90 = 1810

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Posts: 11713
Re: On a particular test whose scores are distributed normally, the 2nd pe  [#permalink]

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Hello from the GMAT Club BumpBot!

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_________________ Re: On a particular test whose scores are distributed normally, the 2nd pe   [#permalink] 11 Jan 2019, 07:53
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