Last visit was: 26 Apr 2024, 02:34 It is currently 26 Apr 2024, 02:34

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.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Kudos
Math Expert
Joined: 02 Sep 2009
Posts: 92929
Own Kudos [?]: 619099 [3]
Given Kudos: 81609
Send PM
examPAL Representative
Joined: 07 Dec 2017
Posts: 1050
Own Kudos [?]: 1777 [0]
Given Kudos: 26
Send PM
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3726
Own Kudos [?]: 16843 [0]
Given Kudos: 165
Send PM
avatar
Intern
Intern
Joined: 24 Oct 2018
Posts: 8
Own Kudos [?]: 0 [0]
Given Kudos: 0
Send PM
Re: Principal of a school recorded the number of students in each of 15 cl [#permalink]
The standard deviation of the number of students in the 15 classes

Given that the average number of students for all the classes = 30

But, we do not have any information about the number of students in each class.

Therefore, Statement 1 alone is not sufficient to answer this question.



Therefore, Statement 2 alone is sufficient to answer this question
Option B
Intern
Intern
Joined: 25 Jul 2018
Posts: 10
Own Kudos [?]: 7 [0]
Given Kudos: 107
Send PM
Re: Principal of a school recorded the number of students in each of 15 cl [#permalink]
Q: std. deviation between # students in 15 classes

S1: avg
S2: each class same # students

S1: insufficient --> average is not enough to determine std. Says nothing about dispersion.

S2: each class same # students --> so difference of them from median 0 , # students class A = # students each other class = avg. the same too , so data points do not deviate from mean = Standard deviation 0


Hence, B imo.
Intern
Intern
Joined: 21 Apr 2018
Posts: 13
Own Kudos [?]: 2 [0]
Given Kudos: 0
Send PM
Re: Principal of a school recorded the number of students in each of 15 cl [#permalink]
If the number of students in each class is the same; there is no deviation between the classes. In other words, zero deviation.

Answer: B
Manager
Manager
Joined: 05 Oct 2014
Posts: 133
Own Kudos [?]: 38 [0]
Given Kudos: 229
Location: India
Concentration: General Management, Strategy
GMAT Date: 07-23-2015
GMAT 1: 580 Q41 V28
GPA: 3.8
WE:Project Management (Energy and Utilities)
Send PM
Re: Principal of a school recorded the number of students in each of 15 cl [#permalink]
DavidTutorexamPAL wrote:
The answer is E.

This question is resolved using the logic of standard deviation - SD is a measure of how spread out an average is. To find it, we need info about the variance of the values being checked, or the values themselves (the number of students in each class). Without that information, finding standard deviation is impossible.



DavidTutorexamPAL, your explanation might be wrong. Request to check please.
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 32678
Own Kudos [?]: 822 [0]
Given Kudos: 0
Send PM
Re: Principal of a school recorded the number of students in each of 15 cl [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
Re: Principal of a school recorded the number of students in each of 15 cl [#permalink]
Moderator:
Math Expert
92929 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne