Question

Teachers are being trained to standardized the scores they give to students' essays. The same essay was scored by 10 different teachers at the beginning and at the end of their training. The results are shown in the tables.

Answer 1

Answer with explanation:

Ques 7)

Score for essay at Beginning of Teachers Training:

76   81   85   79   89    86    84    80    88    79

The mean of the score is calculated as:

$Mean(x')=\dfrac{76+81+85+79+89+86+84+80+88+79}{10}\\\\\\Mean(x')=\dfrac{827}{10}\\\\\\Mean(x')=82.7$

Now, mean absolute deviation is calculated as:

$MAD=\dfrac{\sum{|x-x'|}}{n}$

where n are the number of points.

  x            |x-x'|

 76           6.7

 81            1.7

 85           2.3

 79           3.7

 89           6.3

 86           3.3

 84            1.3

 80            2.7  

 88            5.3

 79            3.7

Hence, MAD is:

$MAD=\dfrac{6.7+1.7+2.3+3.7+6.3+3.3+1.3+2.7+5.3+3.7}{10}\\\\\\MAD=3.7$

Score for essay at End of Teachers Training:

79   82   84    81    77    85     82    80    78     83

The mean of the score is calculated as:

$Mean(x')=\dfrac{79+82+84+81+77+85+82+80+78+83}{10}\\\\\\Mean(x')=\dfrac{811}{10}\\\\\\Mean(x')=81.1$

Now, mean absolute deviation is calculated as:

$MAD=\dfrac{\sum{|x-x'|}}{n}$

where n are the number of points.

  x            |x-x'|

 79           2.1

 82          0.9

 84           2.9

 81           0.1

 77           4.1

 85           3.9

 82           0.9

 80            1.1

 78            3.1

 83            1.9

Hence, MAD is:

$MAD=\dfrac{2.1+0.9+2.9+0.1+4.1+3.9+0.9+1.1+3.1+1.9}{10}\\\\\\MAD=2.1$

Yes, the teacher made progress in standarizing the score. Since the MAD decreases after the end of training.

Ques 8)

If the MAD is 0 than this means that all the scores are same and hence equal to the mean and would result in the deviation equal to 0 and hence the mean absolute deviation is 0.

Answer 2

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