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Sagot :
To determine which conclusion can be made from the data, let's go through the relevant probabilities step-by-step.
1. Calculate the Total for Each Category:
- Employees who leave on time: [tex]\(5 + 81 = 86\)[/tex]
- Employees who leave late: [tex]\(56 + 9 = 65\)[/tex]
- Employees who received a speeding ticket: [tex]\(5 + 56 = 61\)[/tex]
- Employees who did not receive a speeding ticket: [tex]\(81 + 9 = 90\)[/tex]
- Total number of employees surveyed: [tex]\(86 + 65 = 151\)[/tex]
2. Determine the Required Probabilities:
- Probability of receiving a speeding ticket given that the employee regularly leaves for work early or on time:
[tex]\[ P(\text{Ticket | On Time}) = \frac{5}{86} \approx 0.058 \][/tex]
- Probability of receiving a speeding ticket given that the employee regularly leaves for work late:
[tex]\[ P(\text{Ticket | Late}) = \frac{56}{65} \approx 0.862 \][/tex]
- Probability of not receiving a speeding ticket given that the employee regularly leaves for work late:
[tex]\[ P(\text{No Ticket | Late}) = \frac{9}{65} \approx 0.138 \][/tex]
3. Compare the Probabilities:
- We compare [tex]\(P(\text{Ticket | On Time})\)[/tex] with [tex]\(P(\text{No Ticket | Late})\)[/tex]:
[tex]\[ 0.058 < 0.138 \][/tex]
4. Analyze the Conclusions Based on Probabilities:
- Option A: This states that employees who leave late are less likely to receive a speeding ticket, which contradicts the probability of [tex]\(0.862\)[/tex] (employees leaving late getting a ticket) being much higher.
- Option B: This states that [tex]\(P(\text{Ticket | On Time})\)[/tex] is less than [tex]\(P(\text{No Ticket | Late})\)[/tex], i.e., [tex]\(0.058 < 0.138\)[/tex], which is true.
- Option C: If leaving late and receiving a ticket were independent, the probabilities would not show the distinct difference [tex]\(0.862\)[/tex] vs [tex]\(0.138\)[/tex].
- Option D: This states that [tex]\(P(\text{Ticket | Late}) = P(\text{No Ticket | On Time})\)[/tex], which is not true as [tex]\(0.862 \neq 0.138\)[/tex].
Based on these comparisons, the correct conclusion is:
B. The probability of an employee receiving a speeding ticket given that they regularly leave for work early or on time is less than the probability of an employee not receiving a speeding ticket given that they regularly leave for work late.
1. Calculate the Total for Each Category:
- Employees who leave on time: [tex]\(5 + 81 = 86\)[/tex]
- Employees who leave late: [tex]\(56 + 9 = 65\)[/tex]
- Employees who received a speeding ticket: [tex]\(5 + 56 = 61\)[/tex]
- Employees who did not receive a speeding ticket: [tex]\(81 + 9 = 90\)[/tex]
- Total number of employees surveyed: [tex]\(86 + 65 = 151\)[/tex]
2. Determine the Required Probabilities:
- Probability of receiving a speeding ticket given that the employee regularly leaves for work early or on time:
[tex]\[ P(\text{Ticket | On Time}) = \frac{5}{86} \approx 0.058 \][/tex]
- Probability of receiving a speeding ticket given that the employee regularly leaves for work late:
[tex]\[ P(\text{Ticket | Late}) = \frac{56}{65} \approx 0.862 \][/tex]
- Probability of not receiving a speeding ticket given that the employee regularly leaves for work late:
[tex]\[ P(\text{No Ticket | Late}) = \frac{9}{65} \approx 0.138 \][/tex]
3. Compare the Probabilities:
- We compare [tex]\(P(\text{Ticket | On Time})\)[/tex] with [tex]\(P(\text{No Ticket | Late})\)[/tex]:
[tex]\[ 0.058 < 0.138 \][/tex]
4. Analyze the Conclusions Based on Probabilities:
- Option A: This states that employees who leave late are less likely to receive a speeding ticket, which contradicts the probability of [tex]\(0.862\)[/tex] (employees leaving late getting a ticket) being much higher.
- Option B: This states that [tex]\(P(\text{Ticket | On Time})\)[/tex] is less than [tex]\(P(\text{No Ticket | Late})\)[/tex], i.e., [tex]\(0.058 < 0.138\)[/tex], which is true.
- Option C: If leaving late and receiving a ticket were independent, the probabilities would not show the distinct difference [tex]\(0.862\)[/tex] vs [tex]\(0.138\)[/tex].
- Option D: This states that [tex]\(P(\text{Ticket | Late}) = P(\text{No Ticket | On Time})\)[/tex], which is not true as [tex]\(0.862 \neq 0.138\)[/tex].
Based on these comparisons, the correct conclusion is:
B. The probability of an employee receiving a speeding ticket given that they regularly leave for work early or on time is less than the probability of an employee not receiving a speeding ticket given that they regularly leave for work late.
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