Question

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.
which conclusion can be made from the data?
a. 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.
b. 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.
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 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.

Explanation:

Step1: Calculate probability for regular early/on - time

The formula for conditional probability is \(P(A|B)=\frac{P(A\cap B)}{P(B)}\).
For employees who regularly leave early or on - time (\(n = 86\)), the number of those who received a speeding ticket is \(5\). The probability of an employee who regularly leaves early or on - time receiving a speeding ticket is \(P(\text{Speeding}|\text{Early/On - time})=\frac{5}{86}\approx0.058\).
The number of those who did not receive a speeding ticket is \(81\). The probability of an employee who regularly leaves early or on - time not receiving a speeding ticket is \(P(\text{No Speeding}|\text{Early/On - time})=\frac{81}{86}\approx0.942\).

Step2: Calculate probability for regular late

For employees who regularly leave late (\(n = 65\)), the number of those who received a speeding ticket is \(56\). The probability of an employee who regularly leaves late receiving a speeding ticket is \(P(\text{Speeding}|\text{Late})=\frac{56}{65}\approx0.862\).
The number of those who did not receive a speeding ticket is \(9\). The probability of an employee who regularly leaves late not receiving a speeding ticket is \(P(\text{No Speeding}|\text{Late})=\frac{9}{65}\approx0.138\).

Step3: Analyze option A

The probability of an employee receiving a speeding ticket given they regularly leave for work late (\(\frac{56}{65}\approx0.862\)) is not the same as the probability of an employee receiving a speeding ticket given they regularly leave for work early or on - time (\(\frac{5}{86}\approx0.058\)). So option A is wrong.

Step4: Analyze option B

Since \(P(\text{Speeding}|\text{Late})=\frac{56}{65}\approx0.862\) and \(P(\text{Speeding}|\text{Early/On - time})=\frac{5}{86}\approx0.058\), employees who regularly leave for work late are more likely (not less likely) to receive a speeding ticket than those who regularly leave early or on - time. So option B is wrong.

Step5: Analyze option C

Two events \(A\) (leaving late) and \(B\) (getting a speeding ticket) are independent if \(P(A\cap B)=P(A)\times P(B)\). \(P(\text{Late})=\frac{65}{151}\), \(P(\text{Speeding})=\frac{61}{151}\), \(P(\text{Late}\cap\text{Speeding})=\frac{56}{151}\). And \(\frac{65}{151}\times\frac{61}{151}=\frac{65\times61}{151^{2}}=\frac{3965}{22801}\approx0.174
eq\frac{56}{151}\approx0.371\). So they are not independent. Option C is wrong.

Step6: Analyze option D

The probability of an employee not receiving a speeding ticket given they regularly leave for work late is \(P(\text{No Speeding}|\text{Late})=\frac{9}{65}\approx0.138\). The probability of an employee not receiving a speeding ticket given they regularly leave for work early or on - time is \(P(\text{No Speeding}|\text{Early/On - time})=\frac{81}{86}\approx0.942\). Since \(0.138<0.942\), option D is correct.

Answer:

D. The probability of an employee not receiving a speeding ticket given that they regularly leave for work late is less than the probability of an employee not receiving a speeding ticket given that they regularly leave for work early or on time.