Question

The measure of the lifetime of a random sample of 36 tires of a certain brand is normally distributed with an unknown mean, standard deviation of 6 months, and a sample mean of 60 months. Find the 99% confidence interval for the mean.

Answer 1

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:

`\bar{x}\pm z(\frac{s}{\sqrt[]{n}})`

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;

`\begin{gathered} 60\pm2.58(\frac{6}{\sqrt[]{36}}) \\ =60\pm2.58((6)/(6)) \\ =60\pm2.58 \\ \therefore60+2.58=62.58 \\ \text{also, 60-2.58=57.42} \\ \therefore57.42<\mu<62.58 \end{gathered}`

Therefore, the 99% confidence interval for the mean is;

Answer: Option 1