You drive to work 5 days a week for a full year (50 weeks), and with probability

Question

You drive to work 5 days a week for a full year (50 weeks), and with
probability p = 0.02 you get a traffic ticket on any given day, independent of other
days. Let X be the total number of tickets you get in the year.

(c) Any one of the tickets is $10 or $20 or $50 with respective probabilities 0.5, 0.3,
and 0.2, and independent of other tickets. Find the mean and the variance of the
amount of money you pay in traffic tickets during the year. (Please tell me the detail on how to get mean and variance step by steps. Thank you)

(d) Suppose you don't know the probability p of getting a ticket. but you got 5 tickets
during the year, and you estimate p by the sample mean
p = 5/250 = 0.02.

What is the range of possible values of p assuming that the difference between
p and the sample mean p is within 5 times the standard deviation of the sample
mean? (Also provide more details on getting the stadnard deviation, sample mean, and the expected mean. Thank you!)

Explanation / Answer

Here if X is the total number of tickets i get in the year in 250 days. WHich have a binomial distribution

where n = 250 and p = 0.02

Here E(X) = 250 * 0.02 = 5

Var(X) = 250 * 0.02 * 0.98 = 4.9

Similarly, if Y is the amount of ticket.

E(Y) = 10 * 0.5 + 20 * 0.3 + 50 * 0.2 = $ 21

Var(Y) = 0.5 * (10 - 21)2 + 0.3 * (20 - 21)2 + 0.2 * (50 - 21)2 = 229

Here Z is the amount of money we pay in traffic tickets during the year

so, Z = XY (As X and Y are independent here)

E(Z) = E(XY) = E(X) E(Y) = 5 * 21 = $ 105

Var(Z) = Var(XY) = Var(X)Var(Y) + Var(X) E(Y)2 + Var(Y) E(X)2

Var(Z) = 4.9 * 229 + 4.9 * 212 + 229 * 52 = 9008

Standard deviation (Z) = sqrt (9008) = $94.91

(d) Here as we estimate the sample proportion = 0.02

Here sample mean proportion p= 0.02

Standard deviation of proportion p= sqrt (0.02 * 0.98/250) = 0.0089

the range is = p +- 5 * p = 0.02 +- 5 * 0.0089 = (0.00, 0.0643)

So here range of p is from 0 to 0.0643

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