Answer:
0.195
0.191
Step-by-step explanation:
Given that :
μ = 4
Using poisson :
P(x) = (e^-μ * μ^x) ÷ x!
P(x = 3) = (e^-4 * 4^3) / 3!
= (0.0183156 * 64) / 6
= 0.1953664
= 0.195
b. What is the probability that he will sell at most 5 policies in two weeks period?
Mean per week = 4
Mean in two weeks = 4 * 2 = 8
P(x ≤ 5) = p(5) + p(4) +... + p(0)
Using the poisson distribution calculator :
P(x ≤ 5) = 0.19124
= 0.191
Suppose x be a random variable that refers to the number of life insurance plans sold by salespeople using Poisson. Assuming that the average number of insurance policies sold per week is
[tex]\lambda=4[/tex]
we know that,
[tex]\bold{P(X=x) = \frac{e^{-\lambda} \times \lambda^{x}} {x!}}[/tex]
For point a:
Probability which is required:
[tex]P(X=3)\\\\\small P(X=3) = \frac{e^{-4} \times 4^{3}} {3! }\\\\\small P(X=3) =0.1954\\\\[/tex]
For point b:
For the next 2 weeks, the average amount of insurance policies sold:
[tex]\small \lambda = 4\times 2=8[/tex]
Probability which is required:
[tex]P(X\leq 5)[/tex] [tex]\text P(X\leq 5)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)P(X=5)}[/tex]
where
[tex]\small P(X=0) = \frac{e^{-8} \times 8^{0}} {0! } = 0.0003[/tex]
Similarly,
[tex]\small P(X=1) = 0.0027\\\\ \small P(X=2) = 0.0107\\\\ \small P(X=3) = 0.0286\\\\ \small P(X=4) = 0.0573\\\\ \small P(X=5) = 0.0916[/tex]
Therefore,
[tex]\small P(X\leq5) = 0.1912[/tex]
Learn more:
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