# A random sample of 100 automobile owners in the state of Virginia shows that an automobile is driven on average 23,500 kilometers per year with a standard deviation of 3900 kilometers. Construct a 99\% confidence interval for the average number of kilometers an automobile is driven annually in Virginia. What can we assert with 99\% confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year?

A random sample of 100 automobile owners in the state of Virginia shows that an automobile is driven on average 23,500 kilometers per year with a standard deviation of 3900 kilometers.
Assume the distribution of measurements to be approximately normal.
a) Construct a $99\mathrm{%}$ confidence interval for the average number of kilometers an automobile is driven annually in Virginia.
b) What can we assert with $99\mathrm{%}$ confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

a) The $99\mathrm{%}$ confidence interval for the average number of kilometres an automobile is driven is obtained below:
The value of mean is 23,500 kilometres, population standard deviation is 3,900 and sample size $\left(n\right)=100$
Critical value:
From the standard normal distribution table, for the $99\mathrm{%}$ confidence level the critical values is $\left({z}^{\cdot }\right)2.58$.
The confidence interval formula for the population mean is,
$C.I.=\stackrel{―}{x}±{z}^{\cdot }\frac{\sigma }{\sqrt{n}}$
Substitute mean $=23,500$, standard deviation $\left(\sigma \right)$ is 3,900 and sample size $\left(n\right)=100$.
$C.I.=\stackrel{―}{x}±{z}^{\cdot }\frac{\sigma }{\sqrt{n}}$
$=23,500±2.58\frac{3.900}{\sqrt{100}}$
$=23,500±2.58\left(390\right)$
$=23,500±1006.2$

Thus, the $99\mathrm{%}$ confidence interval for the average number of kilometres an automobile is between $22,493.8$ and $24,506.2$.
b) The required formula is obtained below:
$ME={z}_{\frac{\alpha }{2}}\left(\frac{\sigma }{\sqrt{n}}\right)$
Substitute 3,900 for $\sigma$, 100 for n and 2.58 for ${Z}_{\frac{\alpha }{2}}$.
$ME=2.58\left(\frac{3,900}{\sqrt{100}}\right)$
$=2.58\left(\frac{3,900}{10}\right)$
$=2.58\left(390\right)$
$=1006.2$
With $99\mathrm{%}$ confidence that the possible size of our error will not exceed the 1006.2 kilometres.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 38, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}