In a survey conducted to see how long Americans keep their cars, 2000 automobile
ID: 3434076 • Letter: I
Question
In a survey conducted to see how long Americans keep their cars, 2000 automobile owners were asked how long they plan to keep their present cars. The results of the survey follow. Find the probability distribution associated with these data. (Enter your answers to four decimal places.) Years Car Is Kept, x Respondents Probability 0 ? x < 1 60 1 ? x < 3 450 3 ? x < 5 335 5 ? x < 7 335 7 ? x < 10 240 10 ? x 580 What is the probability that an automobile owner selected at random from those surveyed plans to keep his or her present car: (a) Less than five years?(b) 3 years or more?
In a survey conducted to see how long Americans keep their cars, 2000 automobile owners were asked how long they plan to keep their present cars. The results of the survey follow. Find the probability distribution associated with these data. (Enter your answers to four decimal places.) Years Car Is Kept, x Respondents Probability 0 ? x < 1 60 1 ? x < 3 450 3 ? x < 5 335 5 ? x < 7 335 7 ? x < 10 240 10 ? x 580 What is the probability that an automobile owner selected at random from those surveyed plans to keep his or her present car: (a) Less than five years?
(b) 3 years or more?
Years Car Is Kept, x Respondents Probability 0 ? x < 1 60 1 ? x < 3 450 3 ? x < 5 335 5 ? x < 7 335 7 ? x < 10 240 10 ? x 580 (a) Less than five years?
(b) 3 years or more?
Years Car Is Kept, x Respondents Probability 0 ? x < 1 60 1 ? x < 3 450 3 ? x < 5 335 5 ? x < 7 335 7 ? x < 10 240 10 ? x 580
Explanation / Answer
P(0 <X<1) = 60/2000 = 0.03
P(1 <X<3) = 450/2000 = 0.225
P(3 <X<5) = 335/2000 = 0.1675
P(5<X<7) = 335/2000 = 0.1675
P(7<X<10) = 240/2000 = 0.12
P(X>10) = 580/2000 = 0.29
a)
P(X<5) = 0.1675 + 0.225 + 0.03 = 0.4225
b)
P(X>=3) = 1- P(X<3)
= 1 - 0.225 - 0.03
= 0.745
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