Answer to Question #338820 in Statistics and Probability for Easy

Question

Akosua works as the customer service officer at the Agarwal Eye Clinic located in Tesano, a suburb of

the Greater Accra region. The number of requests from male patients for emergency eye treatment that

she receives during any day may be modeled by a poisson distribution with mean of 2.6.Find the

probability that the number of requests from male patients for emergency eye treatment that Akosua

receives on a particular day is;

(i) Exactly 4

(ii)At least 1

(b) Find the probability that the number of requests from male patients for emergency eye

treatment that Akosua receives during a period of 5 days is 15 or fewer

Answer 1

Let $X=$ the number of of requests from male patients: $X\sim Po(\lambda t).$

(a) $\lambda t=2.6(1)=2.6$

(i)

$$P(X=4)=\dfrac{e^{-2.6}(2.6)^4}{4!}=0.141422$$

(ii)

$$P(X\ge 1)=1-P(X=0)$$

$$=1-\dfrac{e^{-2.6}(2.6)^0}{0!}=0.925726$$

(b)

$$\lambda t=2.6(5)=13$$

$$P(X\le 15)=P(X=0)+P(X=1)+P(X=2)$$

$$+P(X=3)+P(X=4)+P(X=5)$$

$$+P(X=6)+P(X=7)+P(X=8)$$

$$+P(X=9)+P(X=10)+P(X=11)$$

$$+P(X=12)+P(X=13)+P(X=14)$$

$$+P(X=15)=\dfrac{e^{-13}(13)^0}{0!}+\dfrac{e^{-13}(13)^1}{1!}$$

$$+\dfrac{e^{-13}(13)^2}{2!}+\dfrac{e^{-13}(13)^3}{3!}+\dfrac{e^{-13}(13)^4}{4!}$$

$$+\dfrac{e^{-13}(13)^5}{5!}+\dfrac{e^{-13}(13)^6}{6!}+\dfrac{e^{-13}(13)^7}{7!}$$

$$+\dfrac{e^{-13}(13)^8}{8!}+\dfrac{e^{-13}(13)^9}{9!}+\dfrac{e^{-13}(13)^{10}}{10!}$$

$$+\dfrac{e^{-13}(13)^{11}}{11!}+\dfrac{e^{-13}(13)^{12}}{12!}+\dfrac{e^{-13}(13)^{13}}{13!}$$

$$+\dfrac{e^{-13}(13)^{14}}{14!}+\dfrac{e^{-13}(13)^{15}}{15!}=0.76361$$