The probability distribution X = number of jobs held during the past year for st

a. P(x=1)=0.39+0.27+0.14+0.05=0.85 c. P(x>=1)=0.85 d. Mean=Sum(x*P(x))=0*0.15+1*0.39+2*0.27+3*0.14+4*0.05=1.55

ID: 3218040 • Letter: T

Question

The probability distribution X = number of jobs held during the past year for students at a college is as follows:

Jobs, (X) 0, 1, 2, 3, 4

P(X), Probability 0.15, 0.39, 0.27, 0.14, 0.05

[a] Find the probability that a randomly selected student held two or fewer jobs during the past year. Answer: ______

[b] Find the probability that a randomly selected student held either one or more jobs during the past year. Answer: ______

[c] Find the probability that a randomly selected student held at least one job during the past year. Answer: ______

[d] What is the mean of X? Answer: ______

[e] What is the standard deviation of X?

a. P(x<=2)=0.27+0.39+0.15=0.81

b. P(x>=1)=0.39+0.27+0.14+0.05=0.85

c. P(x>=1)=0.85

d. Mean=Sum(x*P(x))=0*0.15+1*0.39+2*0.27+3*0.14+4*0.05=1.55

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