(a)
Let X denote the random variable representing Richard's fine payment (in $). Now we are given the following info:
P(X=40) = 0.05
P(X=0) = 1 - 0.05 = 0.95
Now, the mean of fine payment:
For Variance, we first calculate:
Thus, variance:
For skewness we first calculate:
Thus, skewness:
(b)
Let Y denote the random variable representing the fine paid by Richard under the pooling arrangement.
The possible values of Y are 0(if no one gets a parking ticket), 20(when one of them gets a parking ticket) and 40(when both get a parking ticket)
Now, P(Y=0) = P(Richard doesnt get a parking ticket)*P(Peter doesnt get a parking ticket)
= (1-0.05)*(1-0.05)
= 0.9025
P(Y=20) = P(Richard gets a ticket)*P(Peter doesnt get a ticket) + P(Richard doesnt get a ticket)*P(Peter gets a ticket)
= 0.05*(1-0.05) + (1-0.05)*0.05
= 0.05*0.95 + 0.05*0.95
= 0.095
P(Y=40) = P(Richard gets a parking ticket)*P(Peter gets a parking ticket)
= 0.05*0.05
= 0.0025
Thus, the distribution of Richard's fine payment under pooling arrangement is given by:
y |
P(Y=y) |
0 | 0.9025 |
20 | 0.0950 |
40 | 0.0025 |
(c)
We first find the first three moments about origin:
Thus, mean:
We observe that :
Mean(X) = Mean(Y)
Var(X) > Var(Y)
Skew(X) > Skew(Y)
Thus, pooling doesn't have any effect on the mean but reduces the variance and skewness of fine payments.
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