Question

Let X equal the number of typos on a printed page with a mean of 4 typos per page.

Answer 1

Answer:

a) There is a 98.17% probability that a randomly selected page has at least one typo on it.

b) There is a 9.16% probability that a randomly selected page has at most one typo on it.

Step-by-step explanation:

Since we only have the mean, we can solve this problem by a Poisson distribution.

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 $\mu = 4$

(a) What is the probability that a randomly selected page has at least one typo on it?

Thats is $P(X \geq 1)$. Either a number is greater or equal than 1, or it is lesser. The sum of the probabilities must be decimal 1. So:

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

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

In which

$P(X < 1) = P(X = 0)$.

So

$P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183$

$P(X \geq 1) = 1 - P(X < 1) = 1 - 0.0183 = 0.9817$

There is a 98.17% probability that a randomly selected page has at least one typo on it.

(b) What is the probability that a randomly selected page has at most one typo on it?

This is $P = P(X = 0) + P(X = 1)$. So:

$P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183$

$P(X = 1) = \frac{e^{-4}*4^{1}}{(1)!} = 0.0733$

$P = P(X = 0) + P(X = 1) = 0.0183 + 0.0733 = 0.0916$

There is a 9.16% probability that a randomly selected page has at most one typo on it.