Given that the mean life of a television set is 97 months, you can set up that:
[tex]\mu=97[/tex]
You also know that the variance is:
[tex]\sigma^2=169[/tex]
You can find the standard deviation by taking the square root of the variance. Then:
[tex]\sigma=\sqrt{169}=13[/tex]
You need to find:
[tex]P(X<100.9)[/tex]
You need to find the z-score with this formula:
[tex]z=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
Knowing that:
[tex]\bar{X}=100.9[/tex]
You can substitute values into the formula and evaluate:
[tex]z=\frac{100.9-97}{\frac{13}{\sqrt{59}}}\approx2.30[/tex]
You have to find:
[tex]P(z<2.30)[/tex]
Using the Standard Normal Distribution Table, you get:
[tex]P(z<2.30)\approx0.9893[/tex]
Then:
[tex]P(X<100.9)\approx0.9893[/tex]
Hence, the answer is:
[tex]P(X<100.9)\approx0.9893[/tex]
