Answer to Question #333978 in Statistics and Probability for laudii

Question

On average, four students visits the Mathematics and Statistics tutoring centre during a 5-minute period. (a) Calculate the probability that three students visit the centre during a 5- minute period. [3] (b) During a ten-minute period, what is the probability that at least 4 students visit the centre? [5] T

Answer 1

We have a Poisson distribution,

$\lambda=\cfrac{4}{5}=0.8 \text{ visits/minute};\\ P_t(X=k)=\cfrac{(\lambda t)^k\cdot e^{-\lambda t}}{k!}=\cfrac{(0.8t)^k\cdot e^{-0.8t}}{k!}.$

$\text{(a) }P_5(X=3)=\cfrac{(0.8\cdot5)^3\cdot e^{-0.8\cdot5}}{3!}=0.1954;\\ \text{(b) }P_{10}(X\ge4)=1-P_{10}(X<4)=\\ =1-(P_{10}(X=0)+P_{10}(X=1)+\\+P_{10}(X=2)+P_{10}(X=3));\\ P_{10}(X=0)=\cfrac{(0.8\cdot10)^0\cdot e^{-0.8\cdot10}}{0!}=0.00034;\\ P_{10}(X=1)=\cfrac{(0.8\cdot10)^1\cdot e^{-0.8\cdot10}}{1!}=0.00268;\\ P_{10}(X=2)=\cfrac{(0.8\cdot10)^2\cdot e^{-0.8\cdot10}}{2!}=0.01073;\\ P_{10}(X=3)=\cfrac{(0.8\cdot10)^3\cdot e^{-0.8\cdot10}}{3!}=0.02863;\\ P_{10}(X\ge4)=1-(0.00034+0.00268+\\ +0.01073+0.02863)=0.95762.$