My son and I are stuck on this one. Can anyone give some insight to this problem? Thank you.
1 answer:
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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.
The first year, the amount is 40,000
the second year is 40000 + 4.2% of 40000, or 0.042 * 4000, so 40000+(0.042*4000)
common factoring that we get 40000(1 + 0.042), or just 40000(1.042)
in short, the starting amount is 40000, and to get the next term's value you'd use the "common ratio" of 1.042, namely the multiplier of 1.042.
for the third year it'll be 40000(1.042) + (0.042 *40000(1.042) ), again, common factoring that
40000(1.042)(1 + 0.042) or 40000(1.042)(1.042) or 40000(1.042)²
therefore,
the second year is 40000 + 4.2% of 40000, or 0.042 * 4000, so 40000+(0.042*4000)
common factoring that we get 40000(1 + 0.042), or just 40000(1.042)
in short, the starting amount is 40000, and to get the next term's value you'd use the "common ratio" of 1.042, namely the multiplier of 1.042.
for the third year it'll be 40000(1.042) + (0.042 *40000(1.042) ), again, common factoring that
40000(1.042)(1 + 0.042) or 40000(1.042)(1.042) or 40000(1.042)²
therefore,