Suppose that the standard deviation of the tube life of a particular brand of TV picture tube is known to be 500, the population of tube life cannot be assumed to be normally distributed. However, the sample mean of x = 8900 is based on a sample of n = 35. Construct the 95% confidence interval for estimating the population mean.

By the Central Limit Theorem if $n$ is sufficiently large, $\bar{X}$ has approximately a normal distribution with $\mu_{\bar{X}}=\mu$ and $\sigma_{\bar{X}}^2=\sigma^2/n.$

The larger the value of $n,$ the better the approximation.

The Central Limit Theorem can generally be used if $n>30.$

The critical value for $\alpha = 0.05$ is $z_c = z_{1-\alpha/2} = 1.96.$

The corresponding confidence interval is computed as shown below:

$CI=(\bar{X}-z_c\times\dfrac{s}{\sqrt{n}}, \bar{X}+z_c\times\dfrac{s}{\sqrt{n}})$

$=(8900-1.96\times\dfrac{500}{\sqrt{35}}, 8900+1.96\times\dfrac{500}{\sqrt{35}})$

$=(8734.35, 9065.65)$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $8734.35<\mu<9065.65,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval  $(8734.35, 9065.65).$