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The high jumps in inches of the students on two intramural track and field teams are shown below. What is the difference of the means as a multiple of the mean absolute deviations? Round your answer to the nearest tenth.


 

High Jumps for Students on Team 1 (in.)

56, 47, 67, 77, 55, 76, 85, 80, 87, 69, 47, 58


 

High Jumps for Students on Team 2 (in.)

52, 32, 65, 63, 52, 70, 72, 61, 54, 43, 29, 32


 
The difference of the means is about ______________ times the MAD.

DragonSlayer554  Nov 27, 2015

Best Answer 

 #1
avatar+1311 
+10

First find the mean average of the data set.

Add these numbers and divide by 12

56, 47, 67, 77, 55, 76, 85, 80, 87, 69, 47, 58

The mean  = 67.00

Then take each number and subtract 67 from it. If it is negative then just make it positive that is called the absolute value.

56 – 67 = 11

47-67 =  20

 67-67 =  0

77-67 = 10

 55-67 = 12

76-67 =  9

85-67 = 18

 80-67 =  13

87-67 =  20

69-67 =  2

47-67 =  20

58-67 =9

Take the results

11, 20, 0, 10, 12, 9, 18, 13, 20, 2, 20, 9

And find the mean average of this.

It is 12.00 That is the Mean Absolute Deviation.

 

Now you have to do that for the second set of data.

I did this but I not put all the work on here.  

The mean is 52.08. The MAD is 12.08

 

Now subtract the lesser mean from the greater mean and that is 67.00-52.08 =  14.92

Now find the mean of the two MADs 12.00 + 12.08 = 24.08. 24.08/2 = 12.04

 

Then divide by the difference of the means by the mean of the two MADs 14.92/12.04 = 1.24

 

Rounding to the nearest tenth it is 1.2

 

This tells us the means of the two data sets differ by about 1.2 times the variability of the two data sets.

 

This is cool but I not know how to use this in real life yet. 

Dragonlance  Nov 29, 2015
Sort: 

2+0 Answers

 #1
avatar+1311 
+10
Best Answer

First find the mean average of the data set.

Add these numbers and divide by 12

56, 47, 67, 77, 55, 76, 85, 80, 87, 69, 47, 58

The mean  = 67.00

Then take each number and subtract 67 from it. If it is negative then just make it positive that is called the absolute value.

56 – 67 = 11

47-67 =  20

 67-67 =  0

77-67 = 10

 55-67 = 12

76-67 =  9

85-67 = 18

 80-67 =  13

87-67 =  20

69-67 =  2

47-67 =  20

58-67 =9

Take the results

11, 20, 0, 10, 12, 9, 18, 13, 20, 2, 20, 9

And find the mean average of this.

It is 12.00 That is the Mean Absolute Deviation.

 

Now you have to do that for the second set of data.

I did this but I not put all the work on here.  

The mean is 52.08. The MAD is 12.08

 

Now subtract the lesser mean from the greater mean and that is 67.00-52.08 =  14.92

Now find the mean of the two MADs 12.00 + 12.08 = 24.08. 24.08/2 = 12.04

 

Then divide by the difference of the means by the mean of the two MADs 14.92/12.04 = 1.24

 

Rounding to the nearest tenth it is 1.2

 

This tells us the means of the two data sets differ by about 1.2 times the variability of the two data sets.

 

This is cool but I not know how to use this in real life yet. 

Dragonlance  Nov 29, 2015
 #2
avatar+91045 
+5

Thanks Dragonlance, the last bit confused me.  I'll think about your answer  :)

Melody  Nov 29, 2015

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