Answer:
a) 22497.7 < μ< 24502.3
b) With 99% confidence the possible error will not exceed 1002.3
Step-by-step explanation:
Given that:
Mean (μ) = 23500 kilometers per year
Standard deviation (σ) = 3900 kilometers
Confidence level (c) = 99% = 0.99
number of samples (n) = 100
a) α = 1 - c = 1 - 0.99 = 0.01
[tex]\frac{\alpha }{2} =\frac{0.01}{2}=0.005\\ z_{\frac{\alpha }{2}}=z_{0.005}=2.57[/tex]
Using normal distribution table, [tex]z_{0.005[/tex] is the z value of 1 - 0.005 = 0.995 of the area to the right which is 2.57.
The margin of error (e) is given as:
[tex]e= z_{0.005}\frac{\sigma}{\sqrt{n} } = 2.57*\frac{3900}{\sqrt{100} } =1002.3[/tex]
The 99% confidence interval = (μ - e, μ + e) = (23500 - 1002.3, 23500 + 1002.3) = (22497.7, 24502.3)
Confidence interval = 22497.7 < μ< 24502.3
b) With 99% confidence the possible error will not exceed 1002.3
