The population has a normal distribution, so the sample means should have a nearly normal distribution with the same mean and with a standard error equal to the standard deviation divided by the square root of the number of items in the sample.
Get a z score from
CORRECTED 12:40 pm 3/7/22:
z = (mean-mu) / (sigma/sqrt(n))
mean = sample mean
mu = populaiton mean
sigma = population standard deviation
sigma/sqrt(n) = standard error of the mean
here,
z = (4.9-5.4) / [(0.9)/(sqrt(25)]
z = (-0.5)/(0.18) = -2.77
The probability their average would be 4.9 or less is equal to the cumulative normal distribution evaluated at 2.77 standard deviations below the mean, giving 0.0028 or 0.28%.
If z = 0, there would have been a 50% probability the sample mean would be less than the population mean.
If we had a sample standard deviation, s, to work with instead of a known population standard deviation, sigma, we would have used Student's t-test instead.