Respuesta :
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:
[tex]P(X = 5) = \frac{e^{-(4)}. (4)^{5}}{5!}[/tex]
= 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
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:
[tex]P(X = 5) = \frac{e^{-(4)}. (4)^{5}}{5!}[/tex]
= 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
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 distribution
- In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
- [tex]\mu[/tex] is the mean in the given interval.
In this problem, the mean is of 4, hence [tex]\mu = 4[/tex].
Item a:
This is P(X = 5), hence:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 5) = \frac{e^{-4}4^{5}}{(5)!} = 0.1563[/tex]
0.1563 = 15.63% probability that exactly five typographical errors are found on a page.
Item b:
This is:
[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]
Hence:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-4}4^{0}}{(0)!} = 0.0183[/tex]
[tex]P(X = 1) = \frac{e^{-4}4^{1}}{(1)!} = 0.0733[/tex]
[tex]P(X = 2) = \frac{e^{-4}4^{2}}{(2)!} = 0.1465[/tex]
[tex]P(X = 3) = \frac{e^{-4}4^{3}}{(3)!} = 0.1954[/tex]
[tex]P(X = 4) = \frac{e^{-4}4^{4}}{(4)!} = 0.1954[/tex]
[tex]P(X = 5) = \frac{e^{-4}4^{5}}{(5)!} = 0.1563[/tex]
Then:
[tex]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[/tex]
0.7852 = 78.52% probability that at most five typographical errors are found on a page.
Item c:
[tex]P(X > 5) = 1 - P(X \leq 5) = 1 - 0.7852 = 0.2148[/tex]
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 https://brainly.com/question/13971530
