BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

SD

Expert replies
by vipulgoyal » Tue May 21, 2013 9:53 pm
E is a collection of four odd integers and the greatest difference between any two integers in E is 4. The standard deviation of E must be one of how many numbers?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

My take 5
Join the discussion
Source: — Problem Solving |

by Atekihcan » Tue May 21, 2013 10:38 pm
Let the smallest integer be 1.
So, the following sets are possible along with their mean and distance from each integer from the mean...
  • {1, 1, 3, 5} # mean = 2.5 # distances of integers from mean {1.5, 1.5, 0.5, 2.5}
    {1, 3, 3, 5} # mean = 3.0 # distances of integers from mean {2.0, 0.0, 0.0, 2.0}
    {1, 3, 5, 5} # mean = 3.5 # distances of integers from mean {2.5, 0.5, 1.5, 1.5}
    {1, 1, 5, 5} # mean = 3.0 # distances of integers from mean {2.0, 2.0, 2.0, 2.0}
    {1, 1, 1, 5} # mean = 2.0 # distances of integers from mean {1.0, 1.0, 1.0, 3.0}
    {1, 5, 5, 5} # mean = 4.0 # distances of integers from mean {3.0, 1.0, 1.0, 1.0}
As we can see there are two identical pairs of distances of the integers from the mean.
So, unique possible number of standard deviations are 4

Answer : B
Join the discussion

by vipulgoyal » Wed May 22, 2013 12:11 am
shoudnt we consider nagetive options like (-1 -1 3 3)
Join the discussion

by Atekihcan » Wed May 22, 2013 12:20 am
vipulgoyal wrote:shoudnt we consider nagetive options like (-1 -1 3 3)
Yes, we should.
But for that particular example you chose, the answer will be different.
As possible sets are {-1, 3, 3, 3}, {-1, -1, 3, 3}, and {-1, -1, -1, 3}
So, possible number of standard deviations cannot be more than 3.

But if all the elements are positive/negative, the answer will be 4.

So, I think the problem should mention that all elements are either positive or negative, as for elements with mixed signs, the answer will be different.
Join the discussion