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Question
gasoline use a random sample of 64 drivers used on average 744 gallons of gasoline per year. the standard deviation of the population is 35 gallons.
part: 0 / 2
part 1 of 2
(a) find the 99% 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.
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Step1: Identify given values
Sample size $n=64$, sample mean $\bar{x}=744$, population standard deviation $\sigma=15$, confidence level 99%.
Step2: Find z-score for 99% confidence
Z-score $z_{0.005}=2.576$ (from standard normal table).
Step3: Calculate standard error
Standard error $SE=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{64}}=1.875$.
Step4: Compute margin of error
Margin of error $E=z\times SE=2.576\times1.875\approx4.83$.
Step5: Find confidence interval
Lower bound: $744 - 4.83\approx739$; Upper bound: $744 + 4.83\approx749$.
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739 < μ < 749