Answer to Question #201387 in Statistics and Probability for Sabahat Qureshi

Question

RCH tumor incidents followed a 

Poisson distribution with λ = 4 tumors per eye for patients with VHL. Using this model, find the probability 

that in a randomly selected patient with VHL:

(a) There are exactly five occurrences of tumors per eye.

(b) There are more than five occurrences of tumors per eye.

(c) There are fewer than five occurrences of tumors per eye.

(d) There are between five and seven occurrences of tumors per eye, inclusive.

Answer 1

$P(X=x) = \frac{e^{-λ}λ^x}{x!} \\ λ=4$

(a)

$P(X=5) = \frac{e^{-4}4^5}{5!} \\ = 0.1562$

(b)

$P(X>5) = 1 -P(X≤5) \\ = 1 -[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)] \\ = 1 -[\frac{e^{-4}4^0}{0!} + \frac{e^{-4}4^1}{1!} + \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!} +\frac{e^{-4}4^5}{5!}] \\ = 1-e^{-4}[1 + 4 + 8+10.66+10.66 +8.53]\\ = 1 -e^{-4}42.85 \\ = 1 -0.7848 \\ = 0.2152$

(c)

$P(X<5) = P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4) \\ = \frac{e^{-4}4^0}{0!} + \frac{e^{-4}4^1}{1!} + \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!} \\ = e^{-4}(1 + 4 + 8+10.66+10.66) \\ = e^{-4} \times 34.32 \\ = 0.6285$

(d)

$P(5≤X≤7)=P(X=5)+P(X=6)+P(X=7) \\ =\frac{e^{-4}4^5}{5!} + \frac{e^{-4}4^6}{6!} +\frac{e^{-4}4^7}{7!} \\ = e^{-4} \times (8.83+5.68+3.25) \\ = e^{-4} \times17.76 \\ =0.3252$