Assume the TV warranty or replacement times for TV sets are normally distributed with a mean of 9.2 years and a standard deviation of 1.1 years. Find the probability that a randomly selected TV will have a replacement time of less than 6 years.

Firstly, let's draw a picture of the distribution:
Graphically, we want to calculate the blue area. To put it differently, we want to calculate
[tex]P(X<6).[/tex]To do this, we need to find the z-score associated with 6. We can calculate it by
[tex]\begin{gathered} z=\frac{6-\mu}{\sigma}\Rightarrow\begin{cases}\mu=\operatorname{mean} \\ \sigma=\text{standard deviation}\end{cases}, \\ \\ z=\frac{6-\mu}{\sigma}=\frac{6-9.2}{1.1}\approx-2.91. \end{gathered}[/tex]Now, we must check our favorite z-score table and look for the probability associated with the z-score we just found. It's 0.0018.
AnswerThe probability that a randomly selected TV will have a warranty of less than 6 years is 0.0018.