Respuesta :
Answer:
(a) The value of P (X = 7) is 0.1388.
(b) The value of P (X ≥ 3) is 0.9380.
(c) The value of P (2 < X < 7) is 0.5433.
(d) [tex]\mu_{X}=6[/tex]
(e) [tex]\sigma_{X}=2.45[/tex]
Step-by-step explanation:
Let X = number of uranium fission tracks on per cm² surface area of the mineral.
The average number of track per cm² surface area is, λ = 6.
The random variable X follows a Poisson distribution with parameter λ = 6.
The probability mass function of a Poisson distribution is:
[tex]P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, 3...[/tex]
(a)
Compute the value of P (X = 7) as follows:
[tex]P(X=6)=\frac{e^{-6}(6)^{7}}{7!}=\frac{0.0025\times 279936}{5040}=0.1388[/tex]
Thus, the value of P (X = 7) is 0.1388.
(b)
Compute the value of P (X ≥ 3) as follows:
P (X ≥ 3) = 1 - P (X < 3)
= 1 - P (X = 0) - P (X = 1) - P (X = 2)
[tex]=1-\frac{e^{-6}(6)^{0}}{0!}-\frac{e^{-6}(6)^{1}}{1!}-\frac{e^{-6}(6)^{2}}{2!}\\=1-0.00248-0.01487-0.04462\\=0.93803\\\approx0.9380[/tex]
Thus, the value of P (X ≥ 3) is 0.9380.
(c)
Compute the value of P (2 < X < 7) as follows:
P (2 < X < 7) = P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)
[tex]=\frac{e^{-6}(6)^{3}}{3!}+\frac{e^{-6}(6)^{4}}{4!}+\frac{e^{-6}(6)^{5}}{5!}+\frac{e^{-6}(6)^{6}}{6!}\\=0.08924+0.13385+0.16062+0.16062\\=0.54433\\\approx0.5443[/tex]
Thus, the value of P (2 < X < 7) is 0.5433.
(d)
The mean of the Poisson distribution is:
[tex]\mu_{X}=\lambda=6[/tex]
(e)
The standard deviation of the Poisson distribution is:
[tex]\sigma_{X}=\sqrt{\sigma^{2}_{X}}=\sqrt{\lambda}=\sqrt{6}=2.4495\approx2.45[/tex]
