Answer:
a)
=4.63 md=4.55 mo= 1.9 b) Sample Standard Deviation≈ 2.58 Coefficient of Variation=55.72% Sample Range=6.9
Step-by-step explanation:
a)
<u>Mean</u>

For the <u>Median</u>, we have to order the entries. So, ordering it goes:
1.9 1.9 2.3 3.9 5.2 5.7 7.3 8.8
Since we have even entries 
mode
The mode for this data 1.9 1.9 2.3 3.9 5.2 5.7 7.3 8.8 is 1.9
b)
<u>Sample Standard Deviation</u>
Here it is the formula to calculate it:

<u>Coefficient of Variation</u>
CV is the quocient between sample Standard deviation over Mean and it is used to make comparisons.

<u>Range</u>
The difference between the highest and the lowest value of this sample
8.8-1.9=6.9