Respuesta :
Answer:
[tex] P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)[/tex]
And we can find the individual probabilities like this:
[tex]P(X=0) =1.4^0 \frac{e^{-1.4}}{0!}= 0.2466[/tex]
[tex]P(X=1) =1.4^1 \frac{e^{-1.4}}{1!}= 0.3452[/tex]
[tex]P(X=2) =1.4^2 \frac{e^{-1.4}}{2!}= 0.2417[/tex]
[tex]P(X=3) =1.4^3 \frac{e^{-1.4}}{3!}= 0.1128[/tex]
And replacing we got:
[tex] P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)=0.2466+ 0.3452+0.2417+0.1128=0.9463[/tex]
Step-by-step explanation:
Definitions and concepts
The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:
[tex]P(X=x) =\lambda^x \frac{e^{-\lambda}}{x!}[/tex]
And the parameter [tex]\lambda=1.4[/tex] represent the average ocurrence rate per unit of time.
Solution to the problem
For this case the batch would be rejected if we found more than 3 defects, so then the probability of accept the batch would be given by:
[tex] P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)[/tex]
And we can find the individual probabilities like this:
[tex]P(X=0) =1.4^0 \frac{e^{-1.4}}{0!}= 0.2466[/tex]
[tex]P(X=1) =1.4^1 \frac{e^{-1.4}}{1!}= 0.3452[/tex]
[tex]P(X=2) =1.4^2 \frac{e^{-1.4}}{2!}= 0.2417[/tex]
[tex]P(X=3) =1.4^3 \frac{e^{-1.4}}{3!}= 0.1128[/tex]
And replacing we got:
[tex] P(X\leq 3)= P(X=0) +P(X=1)+P(X=2) +P(X=3)=0.2466+ 0.3452+0.2417+0.1128=0.9463[/tex]
