Mathematics, 01.08.2019 23:00 elsham3478

Anewspaper finds that the mean number of typographical errors per page is fourfour. find the probability that (a) exactly fivefive typographical errors are found on a page, (b) at most fivefive typographical errors are found on a page, and (c) more than fivefive typographical errors are found on a page.

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Answer from: AlaskaAirlines

Using the Poisson distribution, it is found that there is a:

a) 0.1563 = 15.63% probability that exactly five typographical errors are found on a page.b) 0.7852 = 78.52% probability that at most five typographical errors are found on a page.c) 0.2148 = 21.48% probability that more than five typographical errors are found on a page.

We have only the mean, hence the Poisson distribution is used to solve this question.

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

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

The parameters are:

x is the number of successese = 2.71828 is the Euler number $\mu$ is the mean in the given interval.

In this problem, the mean is of 4, hence $\mu = 4$ .

Item a:

This is P(X = 5), hence:

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

$P(X = 5) = \frac{e^{-4}4^{5}}{(5)!} = 0.1563$

0.1563 = 15.63% probability that exactly five typographical errors are found on a page.

Item b:

This is:

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

Hence:

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

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

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

$P(X = 2) = \frac{e^{-4}4^{2}}{(2)!} = 0.1465$

$P(X = 3) = \frac{e^{-4}4^{3}}{(3)!} = 0.1954$

$P(X = 4) = \frac{e^{-4}4^{4}}{(4)!} = 0.1954$

$P(X = 5) = \frac{e^{-4}4^{5}}{(5)!} = 0.1563$

Then:

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0183 + 0.0733 + 0.1465 + 0.1954 + 0.1954 + 0.1563 = 0.7852$

0.7852 = 78.52% probability that at most five typographical errors are found on a page.

Item c:

$P(X 5) = 1 - P(X \leq 5) = 1 - 0.7852 = 0.2148$

0.2148 = 21.48% probability that more than five typographical errors are found on a page.

You can learn more about the Poisson distribution at link

Answer from: itzyoboyCj

For this question, we are given a mean/average rate, that is 4 errors per page;
This suggests we need to use the Poisson distribution, so:
Let X be the random variable, the number of errors per page;
X~Po(4) ← X has a Poisson distribution with mean (λ) of 4
a)
What we want is the probability that X = 5, so we use the formula for Poisson:
$P(X = 5) = \frac{e^{-(4)}. (4)^{5}}{5!}$
= 0.156 (to 3 s.f.)
b)
What we want to find is the probability that X ≤ 5, so we have to use the cumulative tables:
P(X ≤ 5) = 0.7851
c)
What we want to find is the probability that X > 5, so we have to do a bit of manipulation and then use the cumulative tables;
The tables give values for ≤, so we cannot directly look up a value for >, thus the manipulation:
P(X > 5) = 1 - P(X ≤ 5)
= 1 - (0.7581)
= 0.2419

Answer from: Quest

i think ur answer is (3,-1)

Answer from: Quest

step-by-step explanation:

7+3=10

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