Answer:
Waiter’s B Large Smoothies showed less variability.
Step-by-step explanation:
First, we have to find the mean for Waiter A’s and Waiter B’s Large Smoothies.
1) Mean
Waiter A
19.1 + 20.1 + 20.9 + 19.6 + 20.9 + 19.5 + 19.2 + 19.4 + 20.3 + 20.9 = 199.9
199.9 / 10 = 19.99
Waiter B
20.1 + 19.6 + 20.0 + 20.5 + 19.8 + 20.0 + 20.1 + 19.7 + 19.9 + 20.4 = 200.1
200.1 / 10 = 20.01
Now, we have to use the Mean Absolute Deviation (MAD) to find which waiter’s smoothies showed less variability.
Waiter A
I19.99 - 19.1I = 0.89
I19.99 - 20.1I = 0.11
I19.99 - 20.9I = 0.91
I19.99 - 19.6I = 0.39
I19.99 - 20.9I = 0.91
I19.99 - 19.5I = 0.49
I19.99 - 19.2I = 0.79
I19.99 - 19.4I = 0.59
I19.99 - 20.3I = 0.31
I19.99 - 20.9I = 0.91
0.89 + 0.11 + 0.91 + 0.39 + 0.91 + 0.49 + 0.79 + 0.59 + 0.31 + 0.91 = 6.3
6.3 / 10 = 0.63
Waiter B
I20.01 - 20.1I = 0.09
I20.01 - 19.6I = 0.41
I20.01 - 20.0I = 0.01
I20.01 - 20.5I = 0.49
I20.01 - 19.8I = 0.21
I20.01 - 20.0I = 0.01
I20.01 - 20.1I = 0.09
I20.01 - 19.7I = 0.4
I20.1 - 19.9I = 0.2
I20.1 - 20.4I = 0.3
0.09 + 0.41 + 0.01 + 0.49 + 0.21 + 0.01 + 0.09 + 0.4 + 0.2 + 0.3 = 2.2
2.2 / 10 = 0.22
Now, we just have to compare those numbers.
0.63 > 0.22
So, Waiter B’s smoothies show less variability.
