5.92 a. Let x denote the number of drug deals on this street on a given night. Note that x is discrete. This

text has covered two discrete distributions, the binomial and the Poisson. The binomial distribution

does not apply here, since there is no fixed number of “trials”. However, the Poisson distribution

might be appropriate.

b. To use the Poisson distribution we would have to assume that the drug deals occur randomly and

independently.

a. The mean number of drug deals per night is three, so for the Poisson distribution for one night, λ =

3. If the residents tape for two nights, then λ = 2 x 3 = 6.

Thus, P(film at least 5 drug deals) = P(x ¿ 5) = 1 – P(x < 5)

= 1 – [P(0) + P(1) + P(2) + P(3) + P(4)] = 1 – (.0025 + .0149 + .0446 + .0892 + .1339) = .7149

b. Part c. shows that two nights of taping are insufficient, since P(x ¿ 5) = .7149 < .90. Try taping for

three nights. Then λ = 3 x 3 = 9. P(x ¿ 5) = 1 – (.0001 + .0011 + .0050 + .0150 + .0337) = .9451.

This exceeds the required probability of .90, so the camera should be rented for three nights.