The number of rescue calls received by a rescue squad in a city follows a Poisson distribution with 2.83 per day. The squad can handle at most four calls a day. a. What is the probability that the squad will be able to handle all the calls on a particular day? b. The squad wants to have at least 95% confidence of being able to handle all the calls received in a day. At least how many calls a day should the squad be prepared for? c. Assuming that the squad can handle at most four calls a day, what is the largest value of that would yield 95% confidence that the squad can handle all calls?

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The number of rescue calls received by a rescue squad in a city follows a Poisson distribution with   2.83 per day. The squad can handle at most four calls a day.   a. What is the probability that the squad will be able to handle all the calls on a particular day?   b. The squad wants to have at least 95% confidence of being able to handle all the calls received in a day. At least how many calls a day should the squad be prepared for?   c. Assuming that the squad can handle at most four calls a day, what is the largest value of that would yield 95% confidence that the squad can handle all calls?

opatovaL · Accepted Answer

Step 1 Given Information Number of Rescue calls received follows a Poisson Distribution with μ=2.83 per day The squad can handle at most four calls a day P.M.F for Poisson distribution is given by P(X=x)=e−μ×μxx! Step 2 a) Probability that the squad will be able to handle all the calls on a particular day is P(X≤4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4) e−2.83×2.8300!+e−2.83×2.8311!+e−2.83×2.8322!+e−2.83×2.8333!+e−2.83×2.8344! =0.059+0.167+0.2229+0.1577 =0.8430 Step 3 b) The squad wants to have at least 95% confidence of being able to handle all the calls received in a day. So, to find the number of calls to satisfy the above condition Probability that the number of calls to have at least 95% confidence is P(X≤x)=∑x=0xe−2.83×2.83xx!≥0.95 Taking x=1 P(X≤1)=0.226 (Using Excel = POISSON.DIST(1,2.83,TRUE)) Taking x=2 P(X≤2)=0.462 (Using Excel = POISSON.DIST(2,2.83,TRUE)) Taking x=3 P(X≤3)=0.685 (Using Excel = POISSON.DIST(3,2.83,TRUE)) Taking x=4 P(X≤4)=0.843 (Using Excel = POISSON.DIST(4,2.83,TRUE)) Taking x=5 P(X≤5)=0.932 (Using Excel = POISSON.DIST(5,2.83,TRUE)) Taking x=6 P(X≤6)=0.97 (Using Excel = POISSON.DIST(6,2.83,TRUE)) The number of calls to home at least 95% confidence is 6 Step 4 c) Assume that the squad can handle at most four calls a day, then finding the largest value of µ that would yield 95% confidence that the squad can handle all calls. P(X≤4)=∑x=0xe−2.83×2.83xx!≥0.95 By taking different values of μ, the minimum value of μ to satisfy the above condition P(X≤4)=∑x=0xe−1.97×1.97xx!=0.95001 The minimum value of μ=1.97

The number of rescue calls received by a rescue squad in a city follows a Poisson distribution with 2.83 per day. The squad can handle at most four calls a day. What is the probability that the squad will be able to handle all the calls on a particular day?

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