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The twelve numbers 3,6,9,12,4,6,8,10,12,14,x,y have a mode of 6 and a mean of 8. Find...

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roshan-rox | Valedictorian

Posted September 28, 2013 at 7:15 PM via web

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The twelve numbers 3,6,9,12,4,6,8,10,12,14,x,y have a mode of 6 and a mean of 8.

Find the values of x and y and the median of the above twelve numbers.

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aruv | High School Teacher | Valedictorian

Posted September 28, 2013 at 7:49 PM (Answer #1)

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The twelve numbers 3,6,9,12,4,6,8,10,12,14,x,y have a

3,4,6,6,8,9,x,y,10,12,12,14

It is given that mode is 6. Therefore  either x=6 or y=6.

Mode=3 x median-2 x mean

6=3 x median - 2 x 8

3 median= 6+16

median=22/3

Mean of these twelve numbers is 8. Therefore

8=(3+6+9+12+4+6+8+10+12+14+x+y)/12

96=84+x+y

x+y=12                   (i)

Then from (i) if  x =6 then y =6.

Thus median=22/3 and x=y=6. 

17/3+y=12

y=19/3

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aruv | High School Teacher | Valedictorian

Posted September 28, 2013 at 7:53 PM (Answer #2)

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The twelve numbers are  3,6,9,12,4,6,8,10,12,14,x,y .

It is given that mode is 6. Therefore  either x=6 or y=6.

Mode=3 * median-2 * mean

6=3 * median - 2 * 8

3 median= 6+16

median=22/3

Mean of these twelve numbers is 8. Therefore

8=(3+6+9+12+4+6+8+10+12+14+x+y)/12

96=84+x+y

x+y=12                   (i)

Then from (i) if  x =6 then y =6.

Thus median=22/3 and x=y=6. 

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steveschoen | College Teacher | (Level 3) Assistant Educator

Posted September 28, 2013 at 10:36 PM (Answer #3)

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The number 3,6,9,12,4,6,8,10,12,14 have a sum of 84.  If 12 numbers have an mean if 8, they have to have a sum of 96.  So, the final 2 numbers have to total 12.  Since the mode has to be 6 with these numbers, and right now the numbers are bi-modal with 6 and 12, then one of these numbers has to be 6.  Given that and their sum as to be 12, both numbers have to be 6.  So, we would have:

3,6,6, 6, 9,12,4,6,8,10,12,14

For the 12 numbers.

For the median, we are looking for the middle of the 12 numbers.  First, we have to put them in numerical order:

3, 4, 6, 6, 6, 6, 8, 9, 10, 12, 12, 14

The median is the middle value of these numbers.  However, if you consider, there isn't one single "middle value".  6 and 8 are the middle values.  In this case, we take the average of the middle values:

(6+8)/2 = 7

So, 7 is the median value for these numbers.

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michuraisin | Student, College Junior | Salutatorian

Posted September 29, 2013 at 4:21 AM (Answer #4)

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Since the mode is 6, 6 needs to be the number that appears the most. In the string of numbers that you gave, both 6 and 12 appear twice, so in order for the mode to be 6, either x or y need to be 6. 

When you add the ten numbers you provided, plus the two variables, then the sum is 84+6+y=90+y. For the twelve numbers to have a mean of 8, the total must be 96 (8x12). Therefore:

90+y=96

y=96-90=6

Both variables are 6!

To find the median, arrange the numbers in numerical order:

3,4,6,6,6,6,8,9,10,12,12,14

Since there are twelves numbers, the median would be the average of the middle two numbers, which are 6 and 8. 

(6+8)/2=7

In conclusion, x and y are both 6 and the median is 7.

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