We know that
• The mean is 98.
,• The standard deviation is 6.
,• The given x-value is 110.
First, we find the z-value using the following formula
[tex]Z=\frac{x-\mu}{\sigma}_{}[/tex]Replacing the given information, we have
[tex]Z=\frac{110-98}{6}=\frac{12}{6}=2_{}[/tex]The z-value or z-score is 2.
Then, we use a z-table to find the probability when P(x<110), or P(z<2).
We obtain a probability of 0.97, which approximates to D.
