Question 1090496
z=(x-mean)/sd
for a (61-62)/3 is -1/3 and (63-62)/3=+(1/3)
probability z is between -1/3 and 1/3=0.2611
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for b, the standard deviation is sigma/sqrt(n)=3/sqrt(25)=0.6
now the z is between -1/.6 and 1/.6 or -5/3 and + 5/3 or 0.9044
It is much higher.  The likelihood of a single woman being outside the range is relatively low, but the likelihood of the mean of 25 of them being outside the range is far less.  Any outlier at one end is likely to be counterbalanced by the same on the other end.