The average age of doctors in a certain hospital is 48.0 years old. suppose the distribution of ages is normal and has a standard deviation of 6.0 years. if 9 doctors are chosen at random for a committee, find the probability that the average age of those doctors is less than 48.8 years. assume that the variable is normally distributed.

True mean = mean (or average)+/- Z*SD/sqrt (sample population)

Now,
True mean has to be less than 48.8 years,
Mean (average) = 48.0 years,
SD = 6.0 years, and 
Sample size (n) = 9 doctors

Making Z the subject of the formula
Z= (True mean - mean)/(SD/sqrt (n))
Substituting,
Z=(48.8-48.0)/(6.0/sqrt (9)) = 0.4
From Normal distribution probabilities table,
At Z= 0.4, P(x<0.4) = 0.6554 0r 65.54%

4should be the last one n=4