Answer:
a) 8.103 g
b) 9.2948
c) 0
Step-by-step explanation:
Given:
Data reported:
9.314 g, 9.215 g, 9.323 g, 8.103 g, 9.278 g, and 9.344 g
Now,
All the values except the 8.103 are above 9
Here the data 8.103 varies very much with respect to the other values
Hence,
a) the data 8.103 should be excluded
b) average value of the mass of the penny = [tex]\frac{9.314 + 9.215 + 9.323 + 9.278 + 9.344 }{5}[/tex]
= 9.2948 g
c) Deviation = Mean - Data
9.2948 - 9.314 = -0.0192
9.2948 - 9.215 = 0.0798
9.2948 - 9.323 = -0.0282
9.2948 - 9.278 = 0.0168
9.2948 - 9.344 = -0.0492
Thus,
Average deviation from mean = tex]\frac{-0.0192 + 0.0798 -0.0282 + 0.0168 -0.0492 }{5}[/tex]
= 0