Since the variable is approximately normally distributed, the distribution of sample means will be approximately normal, with a mean of 34.
be the sample mean age of the employees.
The standard deviation of the sample mean is
The problem is to find P(
) = P(
to standard normal units.]
< 0) + P(0 <
< 1.25) = 0.5000 + 0.3944 = 0.8944 = 89.34%
[Use the standard normal distribution table.]
The probability of sample mean age of the employees is less than 35 years is 89.44%.