Answer :
We are given that the measure of the lifetime of a random sample of 36 tires of a certain brand is normally distributed with an unknown mean, the standard deviation of 6 months, and a sample mean of 60 months.
To find the 99% confidenece interval for we will use the formula below:
[tex]\bar{x}\pm z(\frac{s}{\sqrt[]{n}})[/tex]where x bar is the sample mean, z is the z score for the 99% confidence interval, s is the sample standard deviation and n is the number of elements in the random sample.
Therefore, x bar = 60, s =6 and n = 36. The z score for a 99% interval is 2.58.
We will then substitute all the above parameters into the formula above to derive the 99% confidence interval for the mean.
This can be seen below;
[tex]\begin{gathered} 60\pm2.58(\frac{6}{\sqrt[]{36}}) \\ =60\pm2.58(\frac{6}{6}) \\ =60\pm2.58 \\ \therefore60+2.58=62.58 \\ \text{also, 60-2.58=57.42} \\ \therefore57.42<\mu<62.58 \end{gathered}[/tex]Therefore, the 99% confidence interval for the mean is;
Answer: Option 1