Answer :
To determine if watching television and whether it is a workday are approximately independent, we need to evaluate the given probabilities and their relationships.
1. Determine the Marginal Probabilities:
The table provides the relative frequencies for different scenarios:
[tex]\[ \begin{array}{|c|c|c|} \hline & \text{Watch TV} & \text{Do Not Watch TV} \\ \hline \text{Workday} & 0.52 & 0.11 \\ \hline \text{Not a Workday} & 0.35 & 0.02 \\ \hline \end{array} \][/tex]
Marginal Probabilities for Workday and Not a Workday:
[tex]\[ P(\text{Workday}) = 0.52 + 0.11 = 0.63 \][/tex]
[tex]\[ P(\text{Not a Workday}) = 0.35 + 0.02 = 0.37 \][/tex]
Marginal Probabilities for Watching TV and Not Watching TV:
[tex]\[ P(\text{Watch TV}) = 0.52 + 0.35 = 0.87 \][/tex]
[tex]\[ P(\text{Do Not Watch TV}) = 0.11 + 0.02 = 0.13 \][/tex]
2. Calculate the Expected Joint Probabilities under Independence:
If Watching TV and Workday were independent, the joint probabilities would be:
[tex]\[ P(\text{Watch TV and Workday})_{\text{ind}} = P(\text{Watch TV}) \times P(\text{Workday}) = 0.87 \times 0.63 = 0.5481 \][/tex]
[tex]\[ P(\text{Do Not Watch TV and Workday})_{\text{ind}} = P(\text{Do Not Watch TV}) \times P(\text{Workday}) = 0.13 \times 0.63 = 0.0819 \][/tex]
3. Compare the Calculated Joint Probabilities with the Observed Probabilities:
From the table, the observed joint probabilities are:
[tex]\[ P(\text{Watch TV and Workday}) = 0.52 \][/tex]
[tex]\[ P(\text{Do Not Watch TV and Workday}) = 0.11 \][/tex]
Comparison:
[tex]\[ P(\text{Watch TV and Workday})_{\text{ind}} = 0.5481 \neq 0.52 \][/tex]
[tex]\[ P(\text{Do Not Watch TV and Workday})_{\text{ind}} = 0.0819 \neq 0.11 \][/tex]
4. Conclusion:
Since the calculated joint probabilities under the assumption of independence do not match the observed joint probabilities, we conclude that the events are not independent.
Therefore, the events are not independent because [tex]\( 0.5481 \neq 0.52 \)[/tex] and [tex]\( 0.0819 \neq 0.11 \)[/tex].
1. Determine the Marginal Probabilities:
The table provides the relative frequencies for different scenarios:
[tex]\[ \begin{array}{|c|c|c|} \hline & \text{Watch TV} & \text{Do Not Watch TV} \\ \hline \text{Workday} & 0.52 & 0.11 \\ \hline \text{Not a Workday} & 0.35 & 0.02 \\ \hline \end{array} \][/tex]
Marginal Probabilities for Workday and Not a Workday:
[tex]\[ P(\text{Workday}) = 0.52 + 0.11 = 0.63 \][/tex]
[tex]\[ P(\text{Not a Workday}) = 0.35 + 0.02 = 0.37 \][/tex]
Marginal Probabilities for Watching TV and Not Watching TV:
[tex]\[ P(\text{Watch TV}) = 0.52 + 0.35 = 0.87 \][/tex]
[tex]\[ P(\text{Do Not Watch TV}) = 0.11 + 0.02 = 0.13 \][/tex]
2. Calculate the Expected Joint Probabilities under Independence:
If Watching TV and Workday were independent, the joint probabilities would be:
[tex]\[ P(\text{Watch TV and Workday})_{\text{ind}} = P(\text{Watch TV}) \times P(\text{Workday}) = 0.87 \times 0.63 = 0.5481 \][/tex]
[tex]\[ P(\text{Do Not Watch TV and Workday})_{\text{ind}} = P(\text{Do Not Watch TV}) \times P(\text{Workday}) = 0.13 \times 0.63 = 0.0819 \][/tex]
3. Compare the Calculated Joint Probabilities with the Observed Probabilities:
From the table, the observed joint probabilities are:
[tex]\[ P(\text{Watch TV and Workday}) = 0.52 \][/tex]
[tex]\[ P(\text{Do Not Watch TV and Workday}) = 0.11 \][/tex]
Comparison:
[tex]\[ P(\text{Watch TV and Workday})_{\text{ind}} = 0.5481 \neq 0.52 \][/tex]
[tex]\[ P(\text{Do Not Watch TV and Workday})_{\text{ind}} = 0.0819 \neq 0.11 \][/tex]
4. Conclusion:
Since the calculated joint probabilities under the assumption of independence do not match the observed joint probabilities, we conclude that the events are not independent.
Therefore, the events are not independent because [tex]\( 0.5481 \neq 0.52 \)[/tex] and [tex]\( 0.0819 \neq 0.11 \)[/tex].
