5. A rowarcher shad ing raction time of drivers states that, " $$ A 05 \% $$ confldence interval for the mean time it bles fir a diver to apply the brakes afler secing the brake lights on a vehfele in fiont of him is 1.2 to 1.8 moves. " What are the values of the point estimate and margin of error for this interval? a) point carimate $$ =1.5 $$ scconds; margin of errur $$ =95 \% $$. P point estimate $$ =1.2 $$ seconds; margin of error $$ =0.3 $$ seconds. e' Noint estimate $$ =1.5 $$ seconds; marsin of error $$ =0.6 $$ seconds. d. point extimate $$ =1.5 $$ seculdsi mantin of error $$ =0.3 $$ seconds e. pount estimate $$ =1.2 $$ seconds; margin of error $$ =0.6 $$ scconds.

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**Step 1. Identify the given interval**

The confidence interval is given as from $$1.2$$ seconds to $$1.8$$ seconds.

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**Step 2. Compute the point estimate (center of the interval)**

The point estimate is the midpoint of the interval. It is calculated by

$$
\text{Point estimate} = \frac{1.2 + 1.8}{2} = \frac{3.0}{2} = 1.5 \text{ seconds.}
$$

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**Step 3. Compute the margin of error**

The margin of error is half the length of the interval. First, calculate the length of the interval:

$$
\text{Interval Length} = 1.8 - 1.2 = 0.6 \text{ seconds.}
$$

Then, the margin of error is

$$
\text{Margin of Error} = \frac{0.6}{2} = 0.3 \text{ seconds.}
$$

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**Conclusion**

The point estimate is $$1.5$$ seconds and the margin of error is $$0.3$$ seconds. This corresponds to option (d).
The point estimate is 1.5 seconds and the margin of error is 0.3 seconds.

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