I assume that says [tex]n=110[/tex] and [tex]\bar x = 98.0
[/tex] and [tex]\sigma = 0.74
[/tex].
It doesn't really tell us if this is the standard deviation of the individual samples or of the sample average. Since they're related by a factor of [tex] \sqrt{n} [/tex], around ten, some common sense tells us this is the standard deviation of the individual samples.
So we can conclude our sample mean has a standard deviation
[tex]s = \dfrac{ \sigma }{ \sqrt n } = \dfrac{.74}{ \sqrt{110}} = 0.0705562[/tex]
and that [tex]x=98.6[/tex] is
[tex]z = \dfrac{ x- \bar x}{s} = \dfrac{98.6 - 98.0}{0.0705562 } = 8.5[/tex]
That's a huge z score, 8.5 standard deviations away. 110 is big enough we can just use the normal distribution and not worry about t distributions.
We can conclude the 98.6 is very unlikely to be the true mean or close to the true mean of human temperatures.
