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:

a. What is the probability of selecting a carton and finding no defective pens?
This happens for k=0, so the probability is:

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.



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.

Answer:9400 backpacks
Step-by-step explanation
since the number of packs to be sold in a is represented by x,
selling price for 1 week = 35x
cost price for 1 week = 15x
profit = 35x-15x = 20x
additional cost of production = 11,000
this implies that 20x - 11,000
7800
20x - 11,000
7,800
20x
7800+ 11000
20x
18,800
x
18800/2
x
9,400
at least 9,400 packs have to be sold each week to make a profit of $7800
Answer:
(a) 0.9412
(b) 0.9996 ≈ 1
Step-by-step explanation:
Denote the events a follows:
= a person passes the security system
= a person is a security hazard
Given:

Then,

(a)
Compute the probability that a person passes the security system using the total probability rule as follows:
The total probability rule states that: 
The value of P (P) is:

Thus, the probability that a person passes the security system is 0.9412.
(b)
Compute the probability that a person who passes through the system is without any security problems as follows:

Thus, the probability that a person who passes through the system is without any security problems is approximately 1.
3 is the correct slope of the line:)
I need a picture to answer the question