91Ó°ÊÓ

For each set of years (\(x\)), divide the number of respondents by the total number of respondents (2000) to obtain the probability. \(P(0\leq x<1)\) = \(\frac{60}{2000}\) \(P(1\leq x<3)\) = \(\frac{440}{2000}\) \(P(3\leq x<5)\) = \(\frac{360}{2000}\) \(P(5\leq x<7)\) = \(\frac{340}{2000}\) \(P(7\leq x<10)\) = \(\frac{240}{2000}\) \(P(10\leq x)\) = \(\frac{560}{2000}\) Now compute the probabilities: 1. \(P(0\leq x<1)=0.03\) 2. \(P(1\leq x<3)=0.22\) 3. \(P(3\leq x<5)=0.18\) 4. \(P(5\leq x<7)=0.17\) 5. \(P(7\leq x<10)=0.12\) 6. \(P(10\leq x)=0.28\) The probability distribution is: $$ \begin{array}{cc} \hline x & P(x) \\ \hline 0\leq x<1 & 0.03 \\ \hline 1\leq x<3 & 0.22 \\ \hline 3\leq x<5 & 0.18 \\ \hline 5\leq x<7 & 0.17 \\ \hline 7\leq x<10 & 0.12 \\ \hline 10\leq x & 0.28 \\ \hline \end{array} $$

To find the probability of an automobile owner planning to keep their current car for less than 5 years, we will sum the probabilities for each range of years less than 5: \(P(x<5)\) = \(P(0\leq x<1) + P(1\leq x<3) + P(3\leq x<5)\) \(P(x<5)\) = \(0.03 + 0.22 + 0.18\) \(P(x<5)\) = \(0.43\) The probability that an automobile owner selected at random plans to keep their present car for less than 5 years is 0.43.

To find the probability of an automobile owner planning to keep their current car for 3 years or more, we will sum the probabilities for each range of years greater or equal to 3: \(P(x\geq3)\) = \(P(3\leq x<5) + P(5\leq x<7) + P(7\leq x<10) + P(10\leq x)\) \(P(x\geq3)\) = \(0.18 + 0.17 + 0.12 + 0.28\) \(P(x\geq3)\) = \(0.75\) The probability that an automobile owner selected at random plans to keep their present car for 3 years or more is 0.75.