Given:
Mean= 98
Standard Deviation= 3
Sample = 42
Observed value = 98.6
To determine the probability that the sample mean will be greater than 98.6 gallons, we first note the z-score formula:
[tex]\begin{gathered} z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} \\ where: \\ x=observed\text{ value} \\ \mu=mean \\ \sigma=standard\text{ deviation} \\ n=sample \end{gathered}[/tex]Next, we plug in what we know:
[tex]\begin{gathered} z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} \\ z=\frac{98.6-98}{\frac{3}{\sqrt{42}}} \\ Calculate \\ z=1.2961 \end{gathered}[/tex]Then, we also note that:
[tex]P(x>98.6)=P(z>1.2961)=0.0975[/tex]Therefore, the probability that the sample mean will be greater than 98.6 gallons is: 0.0975