Let's start by computing the z score at 116. The mean of the normal distribution and its standard deviation is already provided on the problem. The z-score can be computed using the equation
where μ is the mean of the normal distribution while σ is the standard deviation of the normal distribution. The value of x for this problem is 116. Hence, the z score is
We want to take a look at the probability of picking a random sample with the upper ceiling equal to 116. No lower limit is considered hence we are considering the lowest random sample here. The normal distribution can be shown as
where the shaded curve is the probability where a random sample can be selected. The value 47.7% is retrieved using the empirical rule of normal distribution wherein if you have a z score that is equal to 2, the probability of getting the random sample at the right-hand side will be 47.7%. No lower limit is given in the problem, hence, we are sure that there is a 50% chance that we can get a sample on the left-hand side of the normal distribution. The total probability that we can get a random sample at most 116 is equal to 50% + 47.7% = 97.7%.