2. A football coach recorded his team’s game scores over a football season. The scores are 21, 45, 21, 14, 21, 28, 24, 14, 24, 28.

(a) Find the mean absolute deviation of the data. Round to the nearest tenth.

(b) Interpret the mean absolute deviation of the data within the context of the problem.

(c) Explain how removing the outlier, 45, affects the mean absolute deviation.

(a) The mean is 24

The Absolute deviation is 8.367

(b) Mean is the number of score per game

(b)  The standard  deviation is the normal expected score difference from the mean

(c) Mean reduce

Standard deviation reduce

Step-by-step explanation:

The data is as follows;

21, 45, 21, 14, 21, 28, 24, 14, 24, 28

The  data with n = 10 is presented as follows

x             (x - μ)       (x - μ)²

21      -3         9

45         21         441

21            -3         9

14        -10        100

21         -3            9

28          4        16

24         0      0

14        -10        100

24         0            0

28    4       16

∑x/n =240  ∑(x - μ)²  =  700

24     ∑(x - μ)²/n = σ² = 70

The mean = 24

The Absolute deviation √70 = 8.367

(b) The mean indicates that within the football season, the average game score per game was 24 and the normal variation from the average was 8.367

(c) Removing the outlier, 45 which ls most equal to two means will reduce the mean or average score per game of the team  to as follows

21 ,  21 , 14,  21,  28,  24,  14,  24,  28

∑x = 195 and μ = (∑x)/n = 195/9 = 21.67

The standard deviation = 4.83

Therefore, removing the outlier 45 which is almost double the mean will reduce the mean by almost μ/n and the standard deviation is also reduced.

1 3 5

step-by-step explanation:

i took the test: )

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40