4.92 is 25 standard deviations to the left of the mean.  Until xl2003, the
Normal distribution function was known to give bad results deep into the
tails.  Anything greater than -3 standard deviations is very small.  For 25
standard deviations, you probably would be safe in assuming a probability of
0.  Also, note that the probability of any single value in a continuous
distribution is considered to be zero.

http://support.microsoft.com/kb/827371/en-us
Description of the NORMDIST function in Excel 2003

Your assumption that the density cannot exceed 1 is incorrect.  To see this
easily, note that almost all of the cumulative area is contained within
mean+/-3*SD, which in your case is a region that is only 0.48 wide; therefore
for that area to approach 1, the height must exceed 1 in at least part of
that region.

Alternately, you can calculate it directly for yourself from the definition
of the normal pdf
 =exp(-(((4.92-4.94)/0.08)^2)/2)/sqrt(2*pi())/0.08
Note that the extra parentheses immediately following the minus sign in the
exp() function are required, because Excel follows a non-standard convention
where unary negation is higher in priority than exponentiation.

Jerry