Question

Car security alarms go off at a mean rate of 4.0 per hour in a large Costco parking lot.

Find the probability that in an hour there will be (Round your answers to 4 decimal places.)

Probability
(a) no alarms
(b) fewer than five alarms
(c) more than seven alarms

Answer 1

Answer on Question#39191 - Math - Statistics and Probability

Car security alarms go off at a mean rate of 4.0 per hour in a large Costco parking lot.

Find the probability that in an hour there will be (Round your answers to 4 decimal places.)

Probability

(a) no alarms

(b) fewer than five alarms

(c) more than seven alarms

Solution

We can assume $\xi$ has a Poisson distribution. The poisson distribution formula is

$$P (k) = P r (\xi = k) = \frac {e ^ {- \lambda_ {\lambda} k}}{k !}, k = 0, 1, 2, \dots$$

where $\mathrm{P}(\mathrm{k})$ is the probability of the event that a random variable $\xi$ takes on the value $\mathrm{k}, e$ is the constant 2.718... , $\lambda$ (usually written as the Greek letter lambda) is the average number of events

(in our case, $\lambda = 4$ )

a) $k = 0$

$$P (0) = \frac {e ^ {- k _ {4} 0}}{0 !} = e ^ {- 4} \approx 0. 0 1 8 3$$

b) $\xi < 5$

$$\begin{array}{l} P (\xi < 5) = | \text {additivity of probability} | = P (\xi = 0) + P (\xi = 1) + P (\xi = 2) + P (\xi = 3) + P (\xi = 4) = \\ = \frac {e ^ {- k _ {4} 0}}{0 !} + \frac {e ^ {- k _ {4} 1}}{1 !} + \frac {e ^ {- k _ {4} 2}}{2 !} + \frac {e ^ {- k _ {4} 3}}{3 !} + \frac {e ^ {- k _ {4} 4}}{4 !} \approx 0. 6 2 8 8 \\ \end{array}$$

c) $\xi > 7$

$$\begin{array}{l} P (\xi > 7) = | \text {probability of the complementary events} | = 1 - \left\{P (\xi = 0) + P (\xi = 1) + P (\xi = 2) + P (\xi = 3) + \right. \\ + P (\xi = 4) + P (\xi = 5) + P (\xi = 6) + P (\xi = 7)) = 0. 0 5 1 1 \\ \end{array}$$