TRY IT 4.1 A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12- hour shift. For a random sample of 50 patients, the following information was obtained. Let X = the number of times a patient rings the nurse during a 12-hour shift. For this exercise, x = 0, 1, 2, 3, 4, 5. P(x) = the probability that X takes on value x. Why is this a discrete probability distribution function (two reasons)? P(x) 4 P(x = 0) 50 8 P(x = 1) = 50 1 16 2 P(x = 2) = 50 14 P(x = 3) = 50 4 6 P(x = 4) = 50 5 P(x = 5) = 50 Table 4.3

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TRY IT 4.1
A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-
hour shift. For a random sample of 50 patients, the following information was obtained. Let X = the number of times a
patient rings the nurse during a 12-hour shift. For this exercise, x = 0, 1, 2, 3, 4, 5. P(x) = the probability that X takes
on value x. Why is this a discrete probability distribution function (two reasons)?
P(x)
4
P(x = 0) =
50
8
1
P(x = 1) =
50
16
2
P(x = 2) =
50
14
P(x = 3) =
50
6
4
P(x = 4) =
50
5
P(x = 5) =
50
Table 4.3
3.
LO
Transcribed Image Text:TRY IT 4.1 A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12- hour shift. For a random sample of 50 patients, the following information was obtained. Let X = the number of times a patient rings the nurse during a 12-hour shift. For this exercise, x = 0, 1, 2, 3, 4, 5. P(x) = the probability that X takes on value x. Why is this a discrete probability distribution function (two reasons)? P(x) 4 P(x = 0) = 50 8 1 P(x = 1) = 50 16 2 P(x = 2) = 50 14 P(x = 3) = 50 6 4 P(x = 4) = 50 5 P(x = 5) = 50 Table 4.3 3. LO
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