In a national survey college students were asked, "How often do you wear a seat belt when riding in a car driven by someone else?" The response frequencies appear in the table to the right. (a) Construct a probability model for seat-belt use by a passenger. (b) Woutd you consider it unusual to find a college student who never wears a seat belt when riding in a car driven by someone else?
\begin{tabular}{lc}
\hline Response & Frequency \\
\hline Never & 141 \\
\hline Rarely & 349 \\
\hline Sometimes & 546 \\
\hline Most of the time & 1339 \\
\hline Always & 2516 \\
\hline
\end{tabular}
(a) Complete the table below.
\begin{tabular}{|c|c|c|}
\hline Response & Probability & \\
\hline Never & & (Round to the nearest thousandth as needed.) \\
\hline Rarely & & (Round to the nearest thousandth as needed.) \\
\hline Sometimes & & (Round to the nearest thousandth as needed.) \\
\hline Most of the time & & (Round to the nearest thousandth as needed.) \\
\hline Always & & (Round to the nearest thousandth as needed.) \\
\hline
\end{tabular}
(b) Would you consider it unusual to find a college student who never wears a seat belt when riding in a car driven by someone else?
A. No, because there were 141 people in the survey who said they never wear their seat belt.
B. Yes, because $P($ never $)<0.05$.
C. Yes, because $0.01
Solution
Step 1 ：First, we need to calculate the total number of responses. This is done by adding up all the frequencies: \(141 + 349 + 546 + 1339 + 2516 = 4891\).
Step 2 ：Next, we calculate the probability of each response by dividing the frequency of each response by the total number of responses. For 'Never', this is \(\frac{141}{4891} = 0.029\). For 'Rarely', this is \(\frac{349}{4891} = 0.071\). For 'Sometimes', this is \(\frac{546}{4891} = 0.112\). For 'Most of the time', this is \(\frac{1339}{4891} = 0.274\). For 'Always', this is \(\frac{2516}{4891} = 0.514\).
Step 3 ：The probability model for seat-belt use by a passenger is as follows: \begin{tabular}{|c|c|} \hline Response & Probability \\ \hline Never & \boxed{0.029} \\ \hline Rarely & \boxed{0.071} \\ \hline Sometimes & \boxed{0.112} \\ \hline Most of the time & \boxed{0.274} \\ \hline Always & \boxed{0.514} \\ \hline \end{tabular}
Step 4 ：Finally, we need to determine whether it would be unusual to find a college student who never wears a seat belt when riding in a car driven by someone else. Since the probability of 'Never' is less than 0.05, we can conclude that it would be unusual. So, the answer is B. Yes, because \(P(\text{never}) < 0.05\).
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Solution
Step 1 ：First, we need to calculate the total number of responses. This is done by adding up all the frequencies: \(141 + 349 + 546 + 1339 + 2516 = 4891\).
Step 2 ：Next, we calculate the probability of each response by dividing the frequency of each response by the total number of responses. For 'Never', this is \(\frac{141}{4891} = 0.029\). For 'Rarely', this is \(\frac{349}{4891} = 0.071\). For 'Sometimes', this is \(\frac{546}{4891} = 0.112\). For 'Most of the time', this is \(\frac{1339}{4891} = 0.274\). For 'Always', this is \(\frac{2516}{4891} = 0.514\).
Step 3 ：The probability model for seat-belt use by a passenger is as follows: \begin{tabular}{|c|c|} \hline Response & Probability \\ \hline Never & \boxed{0.029} \\ \hline Rarely & \boxed{0.071} \\ \hline Sometimes & \boxed{0.112} \\ \hline Most of the time & \boxed{0.274} \\ \hline Always & \boxed{0.514} \\ \hline \end{tabular}
Step 4 ：Finally, we need to determine whether it would be unusual to find a college student who never wears a seat belt when riding in a car driven by someone else. Since the probability of 'Never' is less than 0.05, we can conclude that it would be unusual. So, the answer is B. Yes, because \(P(\text{never}) < 0.05\).
