Answer: -0.36
Step-by-step explanation:
Given: Mean : [tex]\mu=10\text{ minutes}[/tex]
Standard deviation : [tex]\sigma=5\text{ minutes}[/tex]
We know that the formula to calculate the z-score is given by :-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x=8.2 minutes, we have
[tex]z=\dfrac{8.2-10}{5}=0.36[/tex]
Hence, the z-score for this amount of advertising time = -0.36