Answer:
2135 hours of operating life.
Step-by-step explanation:
We need here to make use of the definition of z-score and the values from a standard normal table for, first, to determine the z-score for those light bulbs with an operating life with a probability of 5% or less (which is equivalent to those which have failed by this operating life), to then determine the value in hours for the operating life of such lights bulbs.
We know that a z-score is defined as follows:
[tex] \\ z = \frac{x - \mu}{\sigma}[/tex] (1)
The approximate value of a z-score for a probability of 5% (0.05) can be obtained from a standard normal table and must be below the mean, which implies a negative number that defines an area of 5% of such a distribution.
If we use the cumulative standard normal table, we know that the values for probabilities are from -infinity to a value above the mean (a positive value). However, the value about to find (the z-score) is below the mean, and negative. Then, we need to find a value for z-score whose complement must be 5% (0.05). That value is approximately z-score = 1.65 (P(z=1.65) = 0.95053). Then, the complement is 1 - P(z=1.65) = 1 - 0.95053 = 0.04947), which corresponds to a z-score = -1.65 (due to the symmetry of the normal distribution).
Having this information at hand, we can determine the value of x in equation (1) to finally establish the hours of operations for those light bulbs that operate with a probability of 5% or less.
Thus
[tex] \\ z = \frac{x - \mu}{\sigma}[/tex]
[tex] \\ -1.65 = \frac{x - 2300}{100}[/tex]
[tex] \\ -1.65*100 = x - 2300[/tex]
[tex] \\ -1.65*100 + 2300 = x[/tex]
[tex] \\ x = 2135[/tex]
As a result, theoretically, it takes 2135 hours of operating life for the first 5% of light bulbs to fail. In other words, the operating life is up to 2135 hours for 5% or less of light bulbs for this population.
The shaded area of the graph below represents that number.
