Did high gas prices keep Americans from hitting the road this past summer? In a nationwide survey of adults, one variable measured was how many days vacationers spent driving on th... Did high gas prices keep Americans from hitting the road this past summer? In a nationwide survey of adults, one variable measured was how many days vacationers spent driving on the road on their longest trip. Consider the following (partial) probability distribution for the random variable X = the number of days for the longest car trip. a) Suppose the probability of 7 days is twice as likely as the probability of 8 days. What are the two missing probabilities to complete the distribution for X? b) Find the expected number of days for the longest car trip done last summer, E(X), and interpret. Read more

Steps to Solve

1. Set Up the Probabilities Let the probability of 8 days be denoted as $p$. According to the problem, the probability of 7 days is twice the probability of 8 days, so we denote the probability of 7 days as $2p$.

2. Use the Total Probability The sum of all probabilities in a probability distribution must equal 1. The known probabilities are:

• For 4 days: $0.10$
• For 5 days: $0.20$
• For 6 days: (unknown, let's call it $x$)
• For 7 days: $2p$
• For 8 days: $p$

Thus, we can set up the equation: $$ 0.10 + 0.20 + x + 2p + p = 1 $$

3. Express Known Relationships From the total probability equation, simplify: $$ 0.30 + x + 3p = 1 $$

We can express $x$ in terms of $p$: $$ x = 1 - 0.30 - 3p = 0.70 - 3p $$

4. Solve for Probabilities Using known values Given that there's no information for $x$ yet, we will also utilize the fact that the probabilities must be non-negative. In particular, since $p$ must also sum with $2p$ (the probability for 7 days) and $x$ must remain between 0 and 1, we have:

• $2p \geq 0$
• $p \geq 0$
• $x \geq 0$

5. Find the Missing Probabilities Using $x = 0.70 - 3p \geq 0$, we can solve for $p$: $$ 0.70 - 3p \geq 0 \implies p \leq \frac{0.70}{3} \approx 0.2333 $$

Now, substituting back into the probability equation:

• Suppose $p = \frac{0.10}{3} \implies 2p = \frac{0.20}{3}$

Summarizing: $$ p + 2p + 0.10 + 0.20 + 0.25 = 1 $$

6. Expected Value Calculation The expected value $E(X)$ is computed as: $$ E(X) = \sum (value \times probability) = 4(0.10) + 5(0.20) + 6(x) + 7(2p) + 8(p) $$

Substituting in the probability values we find: $$ E(X) = 4(0.10) + 5(0.20) + 6(0.20) + 7(2p) + 8(p) $$

7. Interpretation Interpret the values to understand the average number of days Americans drove on their longest trip.