Respuesta :
Answer:
a) 0.8088 = 80.88% probability that there are at most 2 typos on a page.
b) 0.0858 = 8.58% probability that there are exactly 10 typos in a 5-page paper.
c) 0.001 = 0.1% probability that there are exactly 2 typos on each page in a 5-page paper.
d) 0.717 = 71.7% probability that there is at least one page with no typos in a 5-page paper.
e) 0.2334 = 23.34% probability that there are exactly two pages with no typos in a 5-page paper.
Step-by-step explanation:
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:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
Binomial distribution:
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The number of typos made by a student follows Poisson distribution with the rate of 1.5 typos per page.
This means that [tex]\mu = 1.5n[/tex], in which n is the number of pages.
(a) Find the probability that there are at most 2 typos on a page.
One page, which means that [tex]\mu = 1.5[/tex]
This is
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
In which
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-1.5}*(1.5)^{0}}{(0)!} = 0.2231[/tex]
[tex]P(X = 1) = \frac{e^{-1.5}*(1.5)^{1}}{(1)!} = 0.3347[/tex]
[tex]P(X = 2) = \frac{e^{-1.5}*(1.5)^{2}}{(2)!} = 0.2510[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2231 + 0.3347 + 0.2510 = 0.8088[/tex]
0.8088 = 80.88% probability that there are at most 2 typos on a page.
(b) Find the probability that there are exactly 10 typos in a 5-page paper.
5 pages, which means that [tex]n = 5, \mu = 5(1.5) = 7.5[/tex].
This is P(X = 10). So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 10) = \frac{e^{-7.5}*(7.5)^{10}}{(10)!} = 0.0858[/tex]
0.0858 = 8.58% probability that there are exactly 10 typos in a 5-page paper.
(c) Find the probability that there are exactly 2 typos on each page in a 5-page paper.
Two typos on a page: 0.2510 probability.
Two typos on each of the 5 pages: (0.251)^5 = 0.001
0.001 = 0.1% probability that there are exactly 2 typos on each page in a 5-page paper.
(d) Find the probability that there is at least one page with no typos in a 5-page paper.
0.2231 probability that a page has no typo, so 1 - 0.2231 = 0.7769 probability that there is at least one typo in a page.
(0.7769)^5 = 0.283 probability that every page has at least one typo.
1 - 0.283 = 0.717 probability that there is at least one page with no typos in a 5-page paper.
(e) Find the probability that there are exactly two pages with no typos in a 5-page paper.
Here, we use the binomial distribution.
0.2231 probability that a page has no typo, so [tex]p = 0.02231[/tex]
5 pages, so [tex]n = 5[/tex]
We want P(X = 2). So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{5,2}.(0.2231)^{2}.(0.7769)^{3} = 0.2334[/tex]
0.2334 = 23.34% probability that there are exactly two pages with no typos in a 5-page paper.
