Respuesta :
Answer:
a) [tex] P(X=5) = \frac{e^{-5} 5^5}{5!}=0.1755[/tex]
[tex] P(X=6) = \frac{e^{-5} 5^6}{6!}=0.1462[/tex]
[tex] P(X=7) = \frac{e^{-5} 5^7}{7!}=0.1044 [/tex]
[tex] P(X=8) = \frac{e^{-5} 5^8}{8!}=0.06528 [/tex]
And adding the values we got [tex] P(5 \leq X \leq 8) =0.4914[/tex]
b) [tex] P(X \geq 8) = 1-P(X<8) = 1-P(X\leq 7)=1-[P(X=0) +....+P(X=7)][/tex]
[tex] P(X=0) = \frac{e^{-5} 5^0}{0!}=0.0067[/tex]
[tex] P(X=1) = \frac{e^{-5} 5^1}{1!}=0.0337[/tex]
[tex] P(X=2) = \frac{e^{-5} 5^2}{2!}=0.0842[/tex]
[tex] P(X=3) = \frac{e^{-5} 5^3}{3!}=0.1403[/tex]
[tex] P(X=4) = \frac{e^{-5} 5^4}{4!}=0.1755[/tex]
[tex] P(X=5) = \frac{e^{-5} 5^5}{5!}=0.1755[/tex]
[tex] P(X=6) = \frac{e^{-5} 5^6}{6!}=0.1462[/tex]
[tex] P(X=7) = \frac{e^{-5} 5^7}{7!}=0.1044 [/tex]
[tex] P(X \geq 8) = 1-P(X<8) = 1-P(X\leq 7)=1-[P(X=0) +....+P(X=7)]= 1-0.8666=0.1334 [/tex]
Step-by-step explanation:
For this case the probability of carrying a defective gene is given by:
[tex] p =\frac{1}{200}= 0.005[/tex]
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=1000, p=0.05)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
And we know that if for a binomial distribution the valueof n is very large and the value of p is very low then we can approximate the dsitribution with a Poisson distribution with the following parameter:
[tex] \mu = np = 1000*0.005= 5[/tex]
And for this case then we can approximate [tex] X \sim Poi (\mu = 5)[/tex]
Part a
For this case we want to find this probability:
[tex] P(5 \leq X \leq 8)[/tex]
And we can use the mass function for the Poisson distribution given by:
[tex] f(x) = \frac{e^{-\mu} \mu^x}{x!}[/tex]
And we can find the individual probabilities like this:
[tex] P(X=5) = \frac{e^{-5} 5^5}{5!}=0.1755[/tex]
[tex] P(X=6) = \frac{e^{-5} 5^6}{6!}=0.1462[/tex]
[tex] P(X=7) = \frac{e^{-5} 5^7}{7!}=0.1044 [/tex]
[tex] P(X=8) = \frac{e^{-5} 5^8}{8!}=0.06528 [/tex]
And adding the values we got [tex] P(5 \leq X \leq 8) =0.4914[/tex]
Part b
For this case we want this probability:
[tex] P(X \geq 8)[/tex]
And we can use the complement rule:
[tex] P(X \geq 8) = 1-P(X<8) = 1-P(X\leq 7)=1-[P(X=0) +....+P(X=7)][/tex]
[tex] P(X=0) = \frac{e^{-5} 5^0}{0!}=0.0067[/tex]
[tex] P(X=1) = \frac{e^{-5} 5^1}{1!}=0.0337[/tex]
[tex] P(X=2) = \frac{e^{-5} 5^2}{2!}=0.0842[/tex]
[tex] P(X=3) = \frac{e^{-5} 5^3}{3!}=0.1403[/tex]
[tex] P(X=4) = \frac{e^{-5} 5^4}{4!}=0.1755[/tex]
[tex] P(X=5) = \frac{e^{-5} 5^5}{5!}=0.1755[/tex]
[tex] P(X=6) = \frac{e^{-5} 5^6}{6!}=0.1462[/tex]
[tex] P(X=7) = \frac{e^{-5} 5^7}{7!}=0.1044 [/tex]
[tex] P(X \geq 8) = 1-P(X<8) = 1-P(X\leq 7)=1-[P(X=0) +....+P(X=7)]= 1-0.8666=0.1334 [/tex]
