Answer:
84.13%
Step-by-step explanation:
Population mean (μ) = 500 hours
Standard deviation (σ) = 50 hours
Assuming a normal distribution, for any given number of hours 'X', the z-score is determined by:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
For X=550
[tex]z=\frac{550-500}{50}\\z=1[/tex]
For a z-score of 1, 'X' corresponds to the 84.13-th percentile of a normal distribution.
Therefore, the probability of that a randomly chosen light bulb lasts less than 550 hours, P(X<550), is 84.13%.