Question #333978

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


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Expert's answer
2022-04-27T02:06:45-0400

We have a Poisson distribution,

λ=45=0.8 visits/minute;Pt(X=k)=(λt)keλtk!=(0.8t)ke0.8tk!.\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!}.


(a) P5(X=3)=(0.85)3e0.853!=0.1954;(b) P10(X4)=1P10(X<4)==1(P10(X=0)+P10(X=1)++P10(X=2)+P10(X=3));P10(X=0)=(0.810)0e0.8100!=0.00034;P10(X=1)=(0.810)1e0.8101!=0.00268;P10(X=2)=(0.810)2e0.8102!=0.01073;P10(X=3)=(0.810)3e0.8103!=0.02863;P10(X4)=1(0.00034+0.00268++0.01073+0.02863)=0.95762.\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.

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