Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
May 2110:00 AM EDT
-11:59 PM EDT
Make the most of your break with the most realistic GMAT™ prep. Take up to $700 off select products.
May 2211:00 AM EDT
-11:59 PM EDT
Take 30% off all study plans with code PICNIC - Expires on Monday, May 25th
May 2512:01 AM EDT
-11:59 PM EDT
Use code MEMORIAL15 for 15% off Target Test Prep GMAT plans and get the only prep course rated 5 stars on GMAT Club. Offer valid 5/25 and 5/26 only.
May 2512:01 AM PDT
-11:59 PM PDT
On the 25th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PST- Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
14 Sep 2013, 20:38
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
69% (01:59) correct
31%
(02:01)
wrong
based on 790
sessions
History
Date
Time
Result
Not Attempted Yet
A researcher investigating mosquito populations recorded a list of estimates of mosquito populations at each of \(n\) different marshland locations in a certain country. The distribution of these estimates was symmetric about the average (arithmetic mean), and 70% of these population estimates were within one standard deviation of the mean. Of these estimates, 75 were more than one standard deviation below the mean. Approximately how many of the estimates were within one standard deviation of the mean?
A) 175
B) 350
C) 375
D) 425
E) 500
A) 175
B) 350
C) 375
D) 425
E) 500
02 Nov 2014, 21:40
Kudos
Bookmarks
carcass
Let M = Arithematic Mean
70% Population Estimate is within 1 Standard Deviation of Mean (1D)
Let this 70% = X
i.e. (M-1D) <= X <= (M+1D)
70% of estimate is within 1 SD of Mean, it implies that 30% of estimates lie in the region on the left side of (M-1D) and on the right side of (M+1D).
Since, the estimates are Symmetric around the mean,
15% of Estimate is on left side of (M-1D), and
15% of Estimate is on right side of (M+1D).
Given, 15% of Estimate on left side of (M-1D) = 75
Therefore, X can be determined by:
15/70 = 75/X
X = 350
Hence, Answer is B
Attachments
nd.png [ 50.25 KiB | Viewed 10562 times ]
14 Sep 2013, 23:53
Kudos
Bookmarks
carcass
If 70% of population is within 1 standard deviation mean then 30% are more than 1 standard deviation below the mean.
Taking symmetric about the mean,
75 are more than 1 sd below the mean therefore these 75 constitute 15% (Taking 15% below the mean and 15% above the mean)
Therefore total population is 500.
Therefore 70% of 500 is 350. who are within 1 standard deviation from the mean.
Hence B
