Question

The probability distribution of x is represented by the following table

x 1 2 3 4 5 6

p(x) 0.04 ?? 0.20 0.41 0.13 0.12

a. What is the value of p(x)=2:

b. What is the value of p(3 ≤ x ≤ 6):

c. If this table represents the number of falls your patients have sustained in the past year, what is the probability that your patient has fallen 5 times?

d. If this table represents the number of falls your patients have sustained in the past year, what is the probability that your patient has fallen 4 or more times?

e. If this table represents the number of falls your patients have sustained in the past year, what is the probability that your patient has fallen more than once, but less than four times?

Answer 1

a)

P(x =2) = 1 - ( 0.04 + 0.2 + 0.41 + 0.13 + 0.12)
= 0.10

b)

P(3 <=x <= 6) = 0.20 + 0.41 + 0.13 + 0.12 = 0.86

c)

P(x = 5) = 0.13

d)

P(x> =4) = 0.41 + 0.13 + 0.12 = 0.66

e)

P(1 < x < 4) = 0.1 + 0.20 = 0.30

Answer 2

a. To find the value of p(x=2), we look at the table, and we see that p(x=2) is missing. However, we know that the sum of all probabilities in a probability distribution must equal 1. Therefore, we can find p(x=2) by subtracting the sum of the other probabilities from 1:

p(x=2) = 1 - (0.04 + 0.20 + 0.41 + 0.13 + 0.12) = 1 - 0.90 = 0.10

b. To find the value of p(3 ≤ x ≤ 6), we add the probabilities for x=3, x=4, x=5, and x=6:

p(3 ≤ x ≤ 6) = p(x=3) + p(x=4) + p(x=5) + p(x=6) = 0.20 + 0.41 + 0.13 + 0.12 = 0.86

c. To find the probability that a patient has fallen 5 times (p(x=5)), we simply look at the corresponding probability in the table:

p(x=5) = 0.12

d. To find the probability that a patient has fallen 4 or more times, we need to add the probabilities for x=4, x=5, and x=6:

p(x ≥ 4) = p(x=4) + p(x=5) + p(x=6) = 0.41 + 0.13 + 0.12 = 0.66

e. To find the probability that a patient has fallen more than once but less than four times, we need to add the probabilities for x=2 and x=3:

p(2 < x < 4) = p(x=2) + p(x=3) = 0.10 + 0.20 = 0.30