Question

The probability that a call received by a certain switchboard will be a wrong number is 0.02. Use the Poisson distribution to approximate the probability that among 150 calls received by the switchboard, there are at least two wrong numbers. Round your answer to four decimal places.

Answer 1

Answer:

0.2008 = 20.08% probability that among 150 calls received by the switchboard, there are at least two wrong numbers.

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 interval.

The probability that a call received by a certain switchboard will be a wrong number is 0.02.

150 calls. So:

$\mu = 150*0.02 = 3$

Use the Poisson distribution to approximate the probability that among 150 calls received by the switchboard, there are at least two wrong numbers.

Either there are less than two calls from wrong numbers, or there are at least two calls from wrong numbers. The sum of the probabilities of these events is 1. So

$P(X < 2) + P(X \geq 2) = 1$

We want to find $P(X \geq 2)$. So

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

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

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0498 + 0.1494 = 0.1992$

Then

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1992 = 0.2008$

0.2008 = 20.08% probability that among 150 calls received by the switchboard, there are at least two wrong numbers.