A survey of 6000 adults found that $52 \%$ say that they would take a ride in a fully self-driving car. A normal distribution may be used to model the sample proportion. At $99 \%$ confidence, the margin of error for this estimate was $1.7 \%$. Find the $99 \%$ confidence interval of the true proportion who said that they would take a ride in a self-driving car. Source: World Economic Forum and Boston Consulting Group. $(0.592,0.604)$ $(0.556,0.604)$ $(0.476,0.564)$ $(0.503,0.537)$

Step 1 ：The question is asking for a confidence interval, which is a range of values that likely contains an unknown population parameter. In this case, the parameter is the proportion of adults who would take a ride in a self-driving car. The confidence interval is calculated using the sample proportion and the margin of error. The sample proportion is the proportion of the sample that has the characteristic of interest, in this case, the proportion of adults who would take a ride in a self-driving car. The margin of error is the maximum expected difference between the true population parameter and a sample estimate.

Step 2 ：The formula for a confidence interval is: \[\text{Confidence Interval} = \text{Sample Proportion} \pm \text{Margin of Error}\]

Step 3 ：Given that the sample proportion is $52 \%$ or $0.52$ and the margin of error is $1.7 \%$ or $0.017$, we can substitute these values into the formula to find the confidence interval.

Step 4 ：Substituting the given values into the formula, we get: \[\text{Confidence Interval} = 0.52 \pm 0.017\]

Step 5 ：Calculating the lower and upper bounds of the confidence interval, we get: \[\text{Lower Bound} = 0.52 - 0.017 = 0.503\] and \[\text{Upper Bound} = 0.52 + 0.017 = 0.537\]

Step 6 ：Final Answer: The $99 \%$ confidence interval of the true proportion who said that they would take a ride in a self-driving car is \(\boxed{(0.503,0.537)}\).