Respuesta :
Answer:
a. P(x = 0 | λ = 1.2) = 0.301
b. P(x ≥ 8 | λ = 1.2) = 0.000
c. P(x > 5 | λ = 1.2) = 0.002
Step-by-step explanation:
If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:
[tex]P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}[/tex]
a. What is the probability of selecting a carton and finding no defective pens?
This happens for k=0, so the probability is:
[tex]P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301[/tex]
b. What is the probability of finding eight or more defective pens in a carton?
This can be calculated as one minus the probablity of having 7 or less defective pens.
[tex]P(k\geq8)=1-P(k<8)=1-\sum_{k=0}^7P(k)=1-\sum_{k=0}^7 \frac{1.2^{k}\cdot e^{-1.2}}{k!}[/tex]
[tex]P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\[/tex]
[tex]P(k<8)=\sum_{k=0}^7P(k)\\\\P(k<8)=0.301+0.361+0.217+0.087+0.026+0.006+0.001+0\approx1\\\\\\P(k\geq8)=1-P(k<8)=1-1=0[/tex]
c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?
We can calculate this as we did the previous question, but for k=5.
[tex]P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002[/tex]
