Answer:
You need to find the x and y intercepts so...
The x intercept is (18,0) since x=18
The y intercept is (0,12) since y=12
Step-by-step explanation:
Cover the y when you are finding the y intercept
Cover the x when you are finding the x intercept
8x/8
144/8
=18
12y/12
144/112
=12
Answer:
x < - 3
Step-by-step explanation:
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:
![P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}](https://tex.z-dn.net/?f=P%28k%29%3D%5Cfrac%7B%5Clambda%5E%7Bk%7De%5E%7B-%5Clambda%7D%7D%7Bk%21%7D%3D%20%5Cfrac%7B1.2%5E%7Bk%7D%5Ccdot%20e%5E%7B-1.2%7D%7D%7Bk%21%7D)
a. What is the probability of selecting a carton and finding no defective pens?
This happens for k=0, so the probability is:
![P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301](https://tex.z-dn.net/?f=P%280%29%3D%5Cfrac%7B1.2%5E%7B0%7D%5Ccdot%20e%5E%7B-1.2%7D%7D%7B0%21%7D%3De%5E%7B-1.2%7D%3D0.301)
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.
![P(k\geq8)=1-P(k](https://tex.z-dn.net/?f=P%28k%5Cgeq8%29%3D1-P%28k%3C8%29%3D1-%5Csum_%7Bk%3D0%7D%5E7P%28k%29%3D1-%5Csum_%7Bk%3D0%7D%5E7%20%5Cfrac%7B1.2%5E%7Bk%7D%5Ccdot%20e%5E%7B-1.2%7D%7D%7Bk%21%7D)
![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\\\\](https://tex.z-dn.net/?f=P%280%29%3D1.2%5E%7B0%7D%20%5Ccdot%20e%5E%7B-1.2%7D%2F0%21%3D1%2A0.3012%2F1%3D0.301%5C%5C%5C%5CP%281%29%3D1.2%5E%7B1%7D%20%5Ccdot%20e%5E%7B-1.2%7D%2F1%21%3D1%2A0.3012%2F1%3D0.361%5C%5C%5C%5CP%282%29%3D1.2%5E%7B2%7D%20%5Ccdot%20e%5E%7B-1.2%7D%2F2%21%3D1%2A0.3012%2F2%3D0.217%5C%5C%5C%5CP%283%29%3D1.2%5E%7B3%7D%20%5Ccdot%20e%5E%7B-1.2%7D%2F3%21%3D2%2A0.3012%2F6%3D0.087%5C%5C%5C%5CP%284%29%3D1.2%5E%7B4%7D%20%5Ccdot%20e%5E%7B-1.2%7D%2F4%21%3D2%2A0.3012%2F24%3D0.026%5C%5C%5C%5CP%285%29%3D1.2%5E%7B5%7D%20%5Ccdot%20e%5E%7B-1.2%7D%2F5%21%3D2%2A0.3012%2F120%3D0.006%5C%5C%5C%5CP%286%29%3D1.2%5E%7B6%7D%20%5Ccdot%20e%5E%7B-1.2%7D%2F6%21%3D3%2A0.3012%2F720%3D0.001%5C%5C%5C%5CP%287%29%3D1.2%5E%7B7%7D%20%5Ccdot%20e%5E%7B-1.2%7D%2F7%21%3D4%2A0.3012%2F5040%3D0%5C%5C%5C%5C)
![P(k](https://tex.z-dn.net/?f=P%28k%3C8%29%3D%5Csum_%7Bk%3D0%7D%5E7P%28k%29%5C%5C%5C%5CP%28k%3C8%29%3D0.301%2B0.361%2B0.217%2B0.087%2B0.026%2B0.006%2B0.001%2B0%5Capprox1%5C%5C%5C%5C%5C%5CP%28k%5Cgeq8%29%3D1-P%28k%3C8%29%3D1-1%3D0)
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.
![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](https://tex.z-dn.net/?f=P%28k%3E5%29%3D1-P%28k%5Cleq5%29%3D1-%5Csum_%7Bk%3D0%7D%5E5P%28k%29%5C%5C%5C%5CP%28k%3E5%29%3D1-%280.301%2B0.361%2B0.217%2B0.087%2B0.026%2B0.006%29%5C%5C%5C%5CP%28k%3E5%29%3D1-0.998%3D0.002)