Answer to Question #252153 in Statistics and Probability for yyy

Question

A shop sells five pieces of shirt every day, from 10am to 10pm.

i) What is the probability of selling more than three shirts today?

ii) What is the probability of selling at most thirty-one shirts from Monday to

Thursday?

iii) What is the probability of selling three to eight shirts for 9 hours?

Answer 1

Let $X=$ the number os shirts sold: $X\sim Po(\lambda t)$

i)

$\lambda t=5$

$$P(X>3)=1-P(X\leq3)$$

$$=1-\dfrac{e^{-5}(5)^0}{0!}-\dfrac{e^{-5}(5)^1}{1!}-\dfrac{e^{-5}(5)^2}{2!}-\dfrac{e^{-5}(5)^3}{3!}$$

$$=0.73497$$

ii)

$\lambda t=20$

$$P(X\leq31)=0.99191$$

Normal Approximation to Poisson Distribution

$\mu=\lambda t=20, \sigma^2=\lambda t=20$

$$P(X\leq31)=P(Z\leq\dfrac{31-20}{\sqrt{20}})\approx P(Z\leq2.45967)$$

$$\approx0.99305$$

iii)

$\lambda t=5(0.9)=4.5$

$$P(3\leq X\leq8)=P(X=3)+P(X=4)$$

$$+P(X=5)+P(X=6)+P(X=7)+P(X=8)$$

$$=\dfrac{e^{-4.5}(4.5)^3}{3!}+\dfrac{e^{-4.5}(4.5)^4}{4!}+\dfrac{e^{-4.5}(4.5)^5}{5!}$$

$$+\dfrac{e^{-4.5}(4.5)^6}{6!}+\dfrac{e^{-4.5}(4.5)^7}{7!}+\dfrac{e^{-4.5}(4.5)^8}{8!}$$

$$=0.16872+0.18981+0.17083$$

$$+0.12812+0.08236+0.04633$$

$$=0.78617$$