Answer: 0.9968
Step-by-step explanation:
Let [tex]\overline{X}[/tex] be the sample mean life of television.
Given: [tex]\mu=138,\ \ \ \sigma^2=324\Rightarrow\ \sigma=\sqrt{324}=18,\ \ \ , n=83[/tex]
The probability that the sample mean would be less than 143.4 months will be :
[tex]P(\overline{X}<143.4)=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{143.4-138}{\dfrac{18}{\sqrt{83}}})\\\\=P(Z<2.733)=0.9969\ \ \ [\text{By p-value table}][/tex]
hence, the required probability = 0.9968
