Lenght of hospital stayus. This distribution representsthe lenght of timne a pat
ID: 2953799 • Letter: L
Question
Lenght of hospital stayus. This distribution representsthe lenght of timne a patient spends in a hospital DAYS FREQUENCY 0-3 2 4-7 15 8-11 8 12-15 6 16+ 9 if a patient is selected, find these probabilities. a. patient spends 3 days or fewer in the hospital b. patient spends fewer than 8 days in the hospital c. the patient spends 16 or more days in the hospital d. the patient spends a maximum of 11 days in thehospital. Lenght of hospital stayus. This distribution representsthe lenght of timne a patient spends in a hospital DAYS FREQUENCY 0-3 2 4-7 15 8-11 8 12-15 6 16+ 9 if a patient is selected, find these probabilities. a. patient spends 3 days or fewer in the hospital b. patient spends fewer than 8 days in the hospital c. the patient spends 16 or more days in the hospital d. the patient spends a maximum of 11 days in thehospital.Explanation / Answer
total number of people: 2+15+8+6+9= 40 people a) 2 people stay 3 days or less in the hospital. The chance to selecta patient who stays 3 days or less in the hospital is then: 2 people / 40 people = 1/20=0.05 b) number of people that stay fewer then 8 days in the hospital = 2+15= 17 people P(fewer then 8 days)=17 people/40 people =17/40 c) 9 people stay longer then 16 days in the hospital P(16 days or more)= 9 people / 40 people = 9/40 d)The number of people that stay maximum 11 days inthe hospital = 2+15+8= 25 people P(maximum 11 days)= 25 people/40 people =25/40=5/8 General lessen. Determine the number of people that stay for acertain period in the hospital and divide this by the total numberto obtain the probability!
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