Question 974217
mean is 98.2;;;;;z=(x-mean)/sd   ;; For a sample, (x-mean)/[sd/sqrt(n)]
sd is 0.62
1.  z=(99-98.2)/0.62=  +1.29  90th percentile (slightly more)
2.  see 1.
3.  No.  I am a retired physician, and this may be found for a variety of reasons.  It can be normal, but in of itself, it is usually ignored.  Statistical and clinical significance are two different entities.
4.z=(-0.22)*sqrt(50)/0.62   the SE is 0.62/sqrt(50).  I inverted to divide and put the sqrt in the numerator.  It is the same thing.  z=-2.52;;  Probability of this being random is 0.006.
5. Yes.  It is more than 4 standard deviations above the mean.  This is highly unlikely to be normal.
6. z=1.645  so 0.62*1.645 + Mean =99.22
(x-mean)/sd =1.645, the 95th percentile. It is +1.02 degrees.
7.It is the same amount in the other direction, - 1.02 degrees or 97.18.  Note that for one person, a low temperature is possible whereas for a sample of 50, even 20% of this decrease would be considered significant.