Answer :
To determine whether the events "watching television" and "a workday" are approximately independent, we need to compare the joint probability of both events occurring with the product of their marginal probabilities.
First, let's define the events:
- [tex]\( A \)[/tex]: Event that an employee watches TV.
- [tex]\( B \)[/tex]: Event that it is a workday.
From the given table, we can observe:
- [tex]\( P(A \cap B) \)[/tex]: Probability that an employee watches TV on a workday.
- [tex]\( P(A \cap B^c) \)[/tex]: Probability that an employee watches TV on a non-workday.
- [tex]\( P(A^c \cap B) \)[/tex]: Probability that an employee does not watch TV on a workday.
- [tex]\( P(A^c \cap B^c) \)[/tex]: Probability that an employee does not watch TV on a non-workday.
The provided relative frequencies are:
- [tex]\( P(A \cap B) = 0.52 \)[/tex]
- [tex]\( P(A \cap B^c) = 0.35 \)[/tex]
- [tex]\( P(A^c \cap B) = 0.11 \)[/tex]
- [tex]\( P(A^c \cap B^c) = 0.02 \)[/tex]
To check for independence, we need to calculate the marginal probabilities:
1. Marginal probability of watching TV ([tex]\( P(A) \)[/tex]):
[tex]\[ P(A) = P(A \cap B) + P(A \cap B^c) = 0.52 + 0.35 = 0.87 \][/tex]
2. Marginal probability of not watching TV ([tex]\( P(A^c) \)[/tex]):
[tex]\[ P(A^c) = P(A^c \cap B) + P(A^c \cap B^c) = 0.11 + 0.02 = 0.13 \][/tex]
3. Marginal probability of a workday ([tex]\( P(B) \)[/tex]):
[tex]\[ P(B) = P(A \cap B) + P(A^c \cap B) = 0.52 + 0.11 = 0.63 \][/tex]
4. Marginal probability of not a workday ([tex]\( P(B^c) \)[/tex]):
[tex]\[ P(B^c) = P(A \cap B^c) + P(A^c \cap B^c) = 0.35 + 0.02 = 0.37 \][/tex]
Next, calculate the expected joint probability if [tex]\( A \)[/tex] and [tex]\( B \)[/tex] were independent:
- Expected joint probability of watching TV and a workday ([tex]\( P(A) \times P(B) \)[/tex]):
[tex]\[ P(A) \times P(B) = 0.87 \times 0.63 \approx 0.5481 \][/tex]
- Expected joint probability of watching TV and not a workday ([tex]\( P(A) \times P(B^c) \)[/tex]):
[tex]\[ P(A) \times P(B^c) = 0.87 \times 0.37 \approx 0.3219 \][/tex]
Compare these with the given joint probabilities:
- Given joint probability of watching TV and a workday ([tex]\( P(A \cap B) \)[/tex]):
[tex]\[ P(A \cap B) = 0.52 \][/tex]
- Expected: 0.5481
- Since [tex]\( 0.52 \neq 0.5481 \)[/tex], it suggests a deviation from independence.
- Given joint probability of watching TV and not a workday ([tex]\( P(A \cap B^c) \)[/tex]):
[tex]\[ P(A \cap B^c) = 0.35 \][/tex]
- Expected: 0.3219
- Since [tex]\( 0.35 \neq 0.3219 \)[/tex], it suggests a deviation from independence.
Since in both cases, the given joint probabilities do not equal the product of their marginal probabilities, we conclude that:
- The events are not independent.
Therefore, the fourth choice is justified:
The events are not independent because [tex]\(0.825 \neq 0.630\)[/tex].
First, let's define the events:
- [tex]\( A \)[/tex]: Event that an employee watches TV.
- [tex]\( B \)[/tex]: Event that it is a workday.
From the given table, we can observe:
- [tex]\( P(A \cap B) \)[/tex]: Probability that an employee watches TV on a workday.
- [tex]\( P(A \cap B^c) \)[/tex]: Probability that an employee watches TV on a non-workday.
- [tex]\( P(A^c \cap B) \)[/tex]: Probability that an employee does not watch TV on a workday.
- [tex]\( P(A^c \cap B^c) \)[/tex]: Probability that an employee does not watch TV on a non-workday.
The provided relative frequencies are:
- [tex]\( P(A \cap B) = 0.52 \)[/tex]
- [tex]\( P(A \cap B^c) = 0.35 \)[/tex]
- [tex]\( P(A^c \cap B) = 0.11 \)[/tex]
- [tex]\( P(A^c \cap B^c) = 0.02 \)[/tex]
To check for independence, we need to calculate the marginal probabilities:
1. Marginal probability of watching TV ([tex]\( P(A) \)[/tex]):
[tex]\[ P(A) = P(A \cap B) + P(A \cap B^c) = 0.52 + 0.35 = 0.87 \][/tex]
2. Marginal probability of not watching TV ([tex]\( P(A^c) \)[/tex]):
[tex]\[ P(A^c) = P(A^c \cap B) + P(A^c \cap B^c) = 0.11 + 0.02 = 0.13 \][/tex]
3. Marginal probability of a workday ([tex]\( P(B) \)[/tex]):
[tex]\[ P(B) = P(A \cap B) + P(A^c \cap B) = 0.52 + 0.11 = 0.63 \][/tex]
4. Marginal probability of not a workday ([tex]\( P(B^c) \)[/tex]):
[tex]\[ P(B^c) = P(A \cap B^c) + P(A^c \cap B^c) = 0.35 + 0.02 = 0.37 \][/tex]
Next, calculate the expected joint probability if [tex]\( A \)[/tex] and [tex]\( B \)[/tex] were independent:
- Expected joint probability of watching TV and a workday ([tex]\( P(A) \times P(B) \)[/tex]):
[tex]\[ P(A) \times P(B) = 0.87 \times 0.63 \approx 0.5481 \][/tex]
- Expected joint probability of watching TV and not a workday ([tex]\( P(A) \times P(B^c) \)[/tex]):
[tex]\[ P(A) \times P(B^c) = 0.87 \times 0.37 \approx 0.3219 \][/tex]
Compare these with the given joint probabilities:
- Given joint probability of watching TV and a workday ([tex]\( P(A \cap B) \)[/tex]):
[tex]\[ P(A \cap B) = 0.52 \][/tex]
- Expected: 0.5481
- Since [tex]\( 0.52 \neq 0.5481 \)[/tex], it suggests a deviation from independence.
- Given joint probability of watching TV and not a workday ([tex]\( P(A \cap B^c) \)[/tex]):
[tex]\[ P(A \cap B^c) = 0.35 \][/tex]
- Expected: 0.3219
- Since [tex]\( 0.35 \neq 0.3219 \)[/tex], it suggests a deviation from independence.
Since in both cases, the given joint probabilities do not equal the product of their marginal probabilities, we conclude that:
- The events are not independent.
Therefore, the fourth choice is justified:
The events are not independent because [tex]\(0.825 \neq 0.630\)[/tex].
