Sasoline Use A random sample of 49 drivers used on average 744 gallons of gasoline per year. The standard deviation of the population is 33 gallons.

Part 1 of 2
(a) Find the $90 \%$ confidence interval of the mean for all drivers. Round intermediate answers to at least three decimal places. Round your final answers to the nearest whole number.
\[
736< \mu< 752
\]

Part: $1 / 2$

Part 2 of 2
(b) If a driver said that he used 804 gallons per year, would you believe that?
(Choose one)
that the driver used 804 gallons per year.
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Step 1 ：Given that the sample mean \(\bar{x}\) is 744 gallons, the standard deviation \(\sigma\) is 33 gallons, and the sample size \(n\) is 49 drivers.

Step 2 ：The formula for the confidence interval is \(CI = \bar{x} \pm Z \times \left(\frac{\sigma}{\sqrt{n}}\right)\), where \(Z\) is the Z-score, which is 1.645 for a 90% confidence interval.

Step 3 ：Substitute the given values into the formula: \(CI = 744 \pm 1.645 \times \left(\frac{33}{\sqrt{49}}\right)\).

Step 4 ：Simplify the expression inside the parentheses: \(CI = 744 \pm 1.645 \times \left(\frac{33}{7}\right)\).

Step 5 ：Calculate the multiplication: \(CI = 744 \pm 1.645 \times 4.714\).

Step 6 ：Calculate the addition and subtraction: \(CI = 744 \pm 7.75\).

Step 7 ：So, the 90% confidence interval for the mean is \(\boxed{736 \text{ to } 752}\) gallons.

Step 8 ：If a driver said that he used 804 gallons per year, it would be outside our confidence interval of 736 to 752 gallons. Therefore, it would be unlikely and we would not readily believe that.