Mathematics: Car security alarms go off at a mean rate of 3.8 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 four alarms (c) more than five alarms

No, Tony is not correct. Solving the inequality tells us that x is greater than or equal to 37.5. Since the class must wash a whole number of cars, they need to wash at least 38 cars. this will help...

Car security alarms go off at a mean rate of - id385212954, ericperkins10ox0b27

Ответ:

gstevens

19.01.2022

Mathematics

a) 0.0224

b) 0.4735

c) 0.1844

Step-by-step explanation:

We use the Poisson probability distribution to solve this problem.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

In this problem, we have that:

$\mu = 3.8$

(a) no alarms

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3.8}*(3.8)^{0}}{(0)!} = 0.0224$

(b) fewer than four alarms

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

$P(X = 0) = \frac{e^{-3.8}*(3.8)^{0}}{(0)!} = 0.0224$

$P(X = 1) = \frac{e^{-3.8}*(3.8)^{1}}{(1)!} = 0.0850$

$P(X = 2) = \frac{e^{-3.8}*(3.8)^{2}}{(2)!} = 0.1615$

$P(X = 3) = \frac{e^{-3.8}*(3.8)^{3}}{(3)!} = 0.2046$

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0224 + 0.0850 + 0.1615 + 0.2046 = 0.4735

Either there are five or fewer alarms, or there are more than five. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X 5) = 1$

We want to find P(X 5) . So

$P(X 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = 0) = \frac{e^{-3.8}*(3.8)^{0}}{(0)!} = 0.0224$

$P(X = 1) = \frac{e^{-3.8}*(3.8)^{1}}{(1)!} = 0.0850$

$P(X = 2) = \frac{e^{-3.8}*(3.8)^{2}}{(2)!} = 0.1615$

$P(X = 3) = \frac{e^{-3.8}*(3.8)^{3}}{(3)!} = 0.2046$

$P(X = 4) = \frac{e^{-3.8}*(3.8)^{4}}{(4)!} = 0.1944$

$P(X = 5) = \frac{e^{-3.8}*(3.8)^{5}}{(5)!} = 0.1477$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0224 + 0.0850 + 0.1615 + 0.2046 + 0.1944 + 0.1477 = 0.8156$

Then

$P(X 5) = 1 - P(X \leq 5) = 1 - 0.8156 = 0.1844$

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Car security alarms go off at a mean rate of 3.8 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 four alarms (c) more than five alarms

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