If the distribution of movie durations is symmetric around the mean of 1.5 hours following a bell curve and if the standard deviation is 20 minutes, approximately what percentage of all movies are longer than 110 minutes?

Question: If the distribution of movie durations is symmetric around the mean of 1.5 hours following a bell curve and if the standard deviation is 20 minutes, approximately what percentage of all movies are longer than 110 minutes?

Solution:
Mean movie duration, µ = 1.5 hours = 90 minutes
Standard deviation, σ = 20 minutes

We are to find the percentage of movies longer than 110 minutes.
So, X = 110
Here, z = (X - µ)/σ
= (110 - 90)/20
= 1

Using standard normal distribution,
Area to the left of z = 1 (i.e. movies ≤ 110 minutes) is approximately 84.13%.
So, the area to the right (i.e., movies > 110 minutes) is = 100% + (- 84.13)% = 15.87% approx 16%