It is currently 23 Aug 2019, 14:51
My Tests

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

On a particular test whose scores are distributed normally,

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
GRE Prep Club Legend
GRE Prep Club Legend
User avatar
Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 128

Kudos [?]: 2046 [0], given: 397

On a particular test whose scores are distributed normally, [#permalink] New post 29 Jul 2018, 04:51
Expert's post
00:00

Question Stats:

40% (01:55) correct 60% (01:13) wrong based on 20 sessions
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
[Reveal] Spoiler: OA

_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Intern
Intern
Joined: 21 Jun 2018
Posts: 35
Followers: 0

Kudos [?]: 6 [0], given: 11

Re: On a particular test whose scores are distributed normally, [#permalink] New post 07 Aug 2018, 22:48
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


Please provide explanation.
3 KUDOS received
Director
Director
Joined: 09 Nov 2018
Posts: 509
Followers: 0

Kudos [?]: 39 [3] , given: 1

Re: On a particular test whose scores are distributed normally, [#permalink] New post 11 Jan 2019, 06:42
3
This post received
KUDOS
ShubhiNigam29 wrote:
Please provide explanation.
Quote:
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


I found the following answer with explanation. I hope this will be helpful.
Attachment:
normdist.png
normdist.png [ 5.65 KiB | Viewed 488 times ]

Answer
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

ANSWER D

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

ANSWER D
Manager
Manager
User avatar
Joined: 22 Jun 2019
Posts: 107
Followers: 0

Kudos [?]: 12 [0], given: 36

CAT Tests
Re: On a particular test whose scores are distributed normally, [#permalink] New post 14 Jul 2019, 04:53
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.

Attachment:
PERCENTILE.png
PERCENTILE.png [ 70.78 KiB | Viewed 183 times ]
Re: On a particular test whose scores are distributed normally,   [#permalink] 14 Jul 2019, 04:53
Display posts from previous: Sort by

On a particular test whose scores are distributed normally,

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

GRE Prep Club Forum Home| About| Terms and Conditions and Privacy Policy| GRE Prep Club Rules| Contact

Powered by phpBB © phpBB Group

Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.