Question

Calculate the absolute deviation (difference) of the mean weight of the kittens and each kitten’s recorded weight.

ITS 50 POINTS TO ANSWER THIS AND IT NEEDS TO BE CORRECT. Sorry its just confusing and I need to pass my class so please help me!

Answer 1

Hi LavenderGamers,

Solution:

Mean - "is the average of the numbers: a calculated "central" value of a set of numbers."

$=\frac{1.25 + 1.5 + 1 + 1.25 + 2 + 0.75 + 2.5 + 1.75 + 2.5 + 1}{10}$

$\frac{15.5}{10} = 1.55$

1.25 - 1.55 = -0.3 = 0.3

1.5 - 1.55 = -0.05 = 0.05

1 - 1.55 = -0.55 = 0.55

1.25 - 1.55 = -0.3 = 0.3

2 - 1.55 = 0.45

0.75 - 1.55 = -0.8 = 0.8

2.5 - 1.55 = 0.95

1.75 - 1.55 = 0.2

2.5 - 1.55 = 0.95

1 - 1.55 = -0.55 = 0.55

Mean Absolute Deviation:

$\frac{0.3 + 0.05 + 0.55 + 0.3 + 0.45 + 0.8 + 0.95 + 0.2 + 0.95 + 0.55}{10}$

$\frac{5.1}{10} = 0.51$

MAD = 0.51

Hope This Helps!

Answer 2

Answer:

Mean - "is the average of the numbers: a calculated "central" value of a set of numbers."

1.25 - 1.55 = -0.3 = 0.3

1.5 - 1.55 = -0.05 = 0.05

1 - 1.55 = -0.55 = 0.55

1.25 - 1.55 = -0.3 = 0.3

2 - 1.55 = 0.45

0.75 - 1.55 = -0.8 = 0.8

2.5 - 1.55 = 0.95

1.75 - 1.55 = 0.2

2.5 - 1.55 = 0.95

1 - 1.55 = -0.55 = 0.55

Mean Absolute Deviation:

MAD = 0.51

Step-by-step explanation: