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The probabilities of the Intensive Care Unit stays, using the exponential distribution, are given as follows:a) One day or less: 0.2368 = 23.68%.b) Between two and three days: 0.1379 = 13.79%.c) More than five days: 0.2589 = 25.89%.Exponential distributionThe exponential probability distribution, with mean given by m, is described by the following equation: [tex]f(x) = \mu e^{-\mu x}[/tex]In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.The cumulative probability distribution function is:[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]Hence:[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]For this problem, the mean and the decay parameter are given, respectively, by:[tex]m = 3.7, \mu = \frac{1}{3.7}[/tex]Hence the probability that an ICU stay is of one day or less is of:[tex]P(X \leq 1) = 1 - e^{-\frac{1}{3.7}} = 0.2368[/tex]The probability that it is between two and three days is obtained as follows:[tex]P(X \leq 2) = 1 - e^{-\frac{2}{3.7}} = 0.4176[/tex][tex]P(X \leq 3) = 1 - e^{-\frac{3}{3.7}} = 0.5555[/tex]0.5555 - 0.4176 = 0.1379.The probability that the stay is more than five days is:[tex]P(X > 5) = e^{-\frac{5}{3.7}} = 0.2589[/tex]More can be learned about the exponential distribution at https://brainly.com/question/14634921#SPJ1