Question

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 1

Answer:

Probability = 0.502

Step-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                2.86

We need to calculate the probability

P(Length of stay of exactly 1 is less than or equal to 4)

P($Y \leq 4$) = P(Y = 1) + P(Y = 2) + P(Y = 3) + P(Y = 4)

$P(Y \leq 4) = 0.0344 + 0.1050 + 0.1546 + 0.2080 = 0.502$

We convert the percent into probabilities by dividing them with 100. This gave us the required probabilities.