Question

select the correct answer.
a survey was given to randomly selected employees who drive to work. each employee was asked to report if they had received a speeding ticket on their morning commute to work any time in the last year. the results are in the table.

	speeding ticket	no speeding ticket	total
regularly leave for work late	56	9	65
total	61	90	151

which conclusion can be made from the data?
a. employees who regularly leave for work late are less likely to receive a speeding ticket than employees who regularly leave for work early or on time.
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.
c. regularly leaving for work late and receiving a speeding ticket are independent events.
d. the probability of an employee receiving a speeding ticket given that they regularly leave for work late is the same as the probability of an employee not receiving a speeding ticket given that they regularly leave for work early or on time.

Explanation:

Step1: Calculate P(ticket|early/on time)

$\frac{5}{86} \approx 0.058$

Step2: Calculate P(ticket|late)

$\frac{56}{65} \approx 0.862$

Step3: Calculate P(no ticket|late)

$\frac{9}{65} \approx 0.138$

Step4: Calculate P(no ticket|early/on time)

$\frac{81}{86} \approx 0.942$

Step5: Check independence for C

$P(\text{ticket}) = \frac{61}{151} \approx 0.404$, $P(\text{ticket}|\text{late})
eq P(\text{ticket})$, so not independent.

Step6: Verify each option

• A: False, late employees are more likely to get tickets.
• B: $0.058 < 0.138$, this is true.
• C: False, events are dependent.
• D: $0.862

eq 0.942$, false.

Answer:

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.