Consider the lengths of stay at a hospital’s emergency department. Hours Count Percent 1 18 3.44 2 55 10.50 3 81 15.46 4 109 20.80 5 88 16.79 6 66 12.60 7 39 7.44 8 17 3.24 9 17 3.24 10 19 3.63 15 15 2.86 Assume that 5 persons independently arrive for service. Round your answers to four decimal places (e.g. 98.7654). (a) What is the probability that the length of stay of exactly 1 person is less than or equal to 4 hours

Answer:Probability = 0.502Step-by-step explanation:We are given the following data :  Hours         Count            Percent   1                    18               3.44   2                    55              10.50   3                    81               15.46   4                    109             20.80   5                     88              16.79   6                     66              12.60   7                     39               7.44   8                     17                3.24   9                     17                3.24   10                   19                3.63    15                  15                2.86We need to calculate the probabilityP(Length of stay of exactly 1 is less than or equal to 4)P([tex]Y \leq 4[/tex]) = P(Y = 1) + P(Y = 2) + P(Y = 3) + P(Y = 4) [tex]P(Y \leq 4) = 0.0344 + 0.1050 + 0.1546 + 0.2080 = 0.502[/tex]We convert the percent into probabilities by dividing them with 100. This gave us the required probabilities.