Question

A type of long-range radio transmits data across the Atlantic Ocean. The number of errors in the transmission during any given amount of time approximately follows a Poisson distribution. The mean number of errors is 2 per hour. (a) What is the probability of having at least 3

Answer 1

Answer:

32.33% probability of having at least 3 erros in an hour.

Step-by-step explanation:

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.

The mean number of errors is 2 per hour.

This means that $\mu = 2$

(a) What is the probability of having at least 3 errors in an hour?

Either you have 2 or less errors in an hour, or we have at least 3 errors. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 2) + P(X \geq 3) = 1$

We want $P(X \geq 3)$

So

$P(X \geq 3) = 1 - P(X \leq 2)$

In which

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

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

$P(X = 0) = \frac{e^{-2}*(2)^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*(2)^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*(2)^{2}}{(2)!} = 0.2707$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1353 + 0.2707 + 0.2707 = 0.6767$

$P(X \geq 3) = 1 - P(X \leq 2) = 1 - 0.6767 = 0.3233$

32.33% probability of having at least 3 erros in an hour.