Answer:
For A: 2.208
For B: 2.9024
Step-by-step explanation:
To solve we will do the following:
For A:
no insured(x) claims(y) xy Var=total/(y-1)
2 1 2 0.08
2 1 2 0.08
1 0 0 0.64
Weighted average 0.8 0.16
a var - X -0.64
k mean/a -1.25
Z (credibility) 1/(1+k) 0.444
no of claims for next year (y=4)
= 0.44*4 + 0.56*0.8
= 2.208
For B:
no insured(x) claims(y) xy Var=total/(y-1)
0 0 0 0
3 6 12 12
2 3 6 2
Weighted average 2.4 3.5
a var - X 1.1
k mean/a 2.18
Z (credibility) 1/(1+k) 0.314
no of claims for next year (y=4)
0.314*4 + 0.686*2.4
= 2.9024
Before the work was done, the cost was ...
($4/gal)*(8 gal)/(340 mi) = $32.00/(340 mi)
After the work was done, the cost was ...
($4/gal)*(7 gal)/(350 mi) = $28.00/(350 mi)
The savings is the difference of these costs:
(32.00/340 -28.00/350) $/mi ≈ $0.01/mi
After the mechanic worked on the car, it cost about $0.01 per mile less to run.
The correct answer is: [C]: " <span>-6.9 < -1.2 " .
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