If age is normally distributed with a mean of 48 and a standard deviation of 18, then you can use the z-score formula and a normal distribution table. The z-score formula is as follows:
z = (x - μ) / σ, where x is the value that you're testing, μ is the mean, and σ is the standard deviation.
So z = (21 - 48) / 18 = -1.5.
Then you look up -1.5 on a standard normal distribution table or use technology to obtain the figure 0.06681, which is 6.681%. This is the percentile that we need because this is the probability that the age is less than 21.