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

Redeem

MEDIAN

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 13
Joined: Mon Apr 18, 2011 4:34 pm

MEDIAN

by SwatiDenre » Thu Jun 16, 2011 8:40 am
The table shows the results of a poll which asked drivers how many accidents they had had over the previous 5 years. What is the median number of accidents per year?

A. 0.5
B. 1
C. 1.5
D. 2
E. 4

Correct Answer: C



Solution ?
Thanks in advance.
Top
Source: — Problem Solving |
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Thu Jun 16, 2011 8:44 am
Hi,
Where is the table, dude...
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
User avatar
Senior | Next Rank: 100 Posts
Posts: 66
Joined: Wed Jul 07, 2010 11:06 am
Thanked: 1 times

by baladon99 » Thu Jun 16, 2011 10:06 am
SwatiDenre wrote:The table shows the results of a poll which asked drivers how many accidents they had had over the previous 5 years. What is the median number of accidents per year?

A. 0.5
B. 1
C. 1.5
D. 2
E. 4

Correct Answer: C



Solution ?
Thanks in advance.
Can you please post the table ?
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 461
Joined: Tue May 10, 2011 9:09 am
Location: pune
Thanked: 36 times
Followed by:3 members

by amit2k9 » Thu Jun 16, 2011 9:32 pm
since the number of years is 5, the median will be the number of accidents in the 3rd year.
For Understanding Sustainability,Green Businesses and Social Entrepreneurship visit -https://aamthoughts.blocked/
(Featured Best Green Site Worldwide-https://bloggers.com/green/popular/page2)
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 143
Joined: Sat Nov 06, 2010 9:06 pm
Thanked: 10 times
Followed by:1 members

by edvhou812 » Thu Jun 16, 2011 10:03 pm
amit2k9 wrote:since the number of years is 5, the median will be the number of accidents in the 3rd year.
No. You will need to reorder the amounts from least to greatest, and then take the third largest number.
Top
Junior | Next Rank: 30 Posts
Posts: 13
Joined: Mon Apr 18, 2011 4:34 pm

by SwatiDenre » Sun Jun 19, 2011 3:51 pm
Hi,
very sorry for the mistake.
here goes the table -


No of accidents - No. of drivers

0 - 17
1 - 13
2 - 21
3 - 4
4 - 2
5 - 2
6 - 1
Top
Master | Next Rank: 500 Posts
Posts: 401
Joined: Tue May 24, 2011 1:14 am
Thanked: 37 times
Followed by:5 members

by MBA.Aspirant » Sun Jun 19, 2011 4:10 pm
The middle of this set will = the number of drivers/2 = 43/2 = 21.5

This number is between two accidents values 1 and 2

so we take the average of them = 1+2/2 = 1.5
Top
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Sun Jun 19, 2011 8:48 pm
SwatiDenre wrote:Hi,
very sorry for the mistake.
here goes the table -


No of accidents - No. of drivers

0 (1-17) - 17
1 (18-30) - 13
2 (31-51) - 21
3 (52-55) - 4
4 (56-57) - 2
5 (58-59) - 2
6 (60) - 1
Hi,
Total number of drivers is 17+13+21+... = 60
Median is the average of 30th and 31st elements
30th element is 1 and 31st element is 2
So, median is (1+2)/2 = 1.5

Hence, C
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
Post new topic Post Reply
• Page 1 of 1