Respuesta :
Answer:
a. The Poisson approximation is good because n is large, p is small, and np < 10.
The parameter of thr Poisson distribution is:
[tex]\lambda =np\approx1.6[/tex]
b. P(r=0)=0.2019
c. P(r>1)=0.4751
d. P(r>2)=0.2167
e. P(r>3)=0.0789
Step-by-step explanation:
a. The Poisson distribution is appropiate to represent binomial events with low probability and many repetitions (small p and large n).
The approximation that the Poisson distribution does to the real model is adequate if the product np is equal or lower than 10.
In this case, n=963 and p=1/596, so we have:
[tex]np=963*(1/596)\approx1.6[/tex]
The Poisson approximation is good because n is large, p is small, and np < 10.
The parameter of thr Poisson distribution is:
[tex]\lambda =np\approx1.6[/tex]
We can calculate the probability for k events as:
[tex]P(r=k)=\dfrac{\lambda^ke^{-\lambda}}{k!}[/tex]
b. P(r=0). We use the formula above with λ=1.6 and r=0.
[tex]P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\[/tex]
c. P(r>1). In this case, is simpler to calculate the complementary probability to P(r<=1), that is the sum of P(r=0) and P(r=1).
[tex]P(r>1)=1-P(r\leq1)=1-[P(r=0)+P(r=1)]\\\\\\P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\P(1)=1.6^{1} \cdot e^{-1.6}/1!=1.6*0.2019/1=0.3230\\\\\\P(r>1)=1-(0.2019+0.3230)=1-0.5249=0.4751[/tex]
d. P(r>2)
[tex]P(r>2)=1-P(r\leq2)=1-[P(r=0)+P(r=1)+P(r=2)]\\\\\\P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\P(1)=1.6^{1} \cdot e^{-1.6}/1!=1.6*0.2019/1=0.3230\\\\P(2)=1.6^{2} \cdot e^{-1.6}/2!=2.56*0.2019/2=0.2584\\\\\\P(r>2)=1-(0.2019+0.3230+0.2584)=1-0.7833=0.2167[/tex]
e. P(r>3)
[tex]P(r>3)=1-P(r\leq2)=1-[P(r=0)+P(r=1)+P(r=2)+P(r=3)]\\\\\\P(0)=1.6^{0} \cdot e^{-1.6}/0!=1*0.2019/1=0.2019\\\\P(1)=1.6^{1} \cdot e^{-1.6}/1!=1.6*0.2019/1=0.3230\\\\P(2)=1.6^{2} \cdot e^{-1.6}/2!=2.56*0.2019/2=0.2584\\\\P(3)=1.6^{3} \cdot e^{-1.6}/3!=4.096*0.2019/6=0.1378\\\\\\P(r>3)=1-(0.2019+0.3230+0.2584+0.1378)=1-0.9211=0.0789[/tex]
