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:
[tex]Mean(x')=\dfrac{76+81+85+79+89+86+84+80+88+79}{10}\\\\\\Mean(x')=\dfrac{827}{10}\\\\\\Mean(x')=82.7[/tex]
Now, mean absolute deviation is calculated as:
[tex]MAD=\dfrac{\sum{|x-x'|}}{n}[/tex]
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:
[tex]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[/tex]
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:
[tex]Mean(x')=\dfrac{79+82+84+81+77+85+82+80+78+83}{10}\\\\\\Mean(x')=\dfrac{811}{10}\\\\\\Mean(x')=81.1[/tex]
Now, mean absolute deviation is calculated as:
[tex]MAD=\dfrac{\sum{|x-x'|}}{n}[/tex]
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:
[tex]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[/tex]
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.
