Information reference tables the lifespan of a 60-watt lightbulb produced by a company is normally distributed with a mean of 1450 hours and a standard deviation of 8.5 hours. if a 60-watt lightbulb produced by this company is selected at random, what is the probability that its lifespan will be between 1440 and 1465 hours

Respuesta :

First, we need to standardize the value 1440 and 1465 to be able to work out the z-score

[tex]z= \frac{1440-1450}{8.5} [/tex]
[tex]z=-1.77[/tex]

[tex]z= \frac{1465-1450}{8.5} [/tex]
[tex]z=1.77[/tex]

The z-scores are shown in the diagram below

To work out the probability of [tex]-1.77\ \textless \ z\ \textless \ 177[/tex] we can first read on the table, the probability when [tex]P(z\ \textless \ 1.77)=0.9616[/tex], then we subtract this value from 1, as we are interested in the area to the right of 1.77.

[tex]1-0.9616=0.0384[/tex]

Then the area between [tex]z=-1.77[/tex] and [tex]z=1.77[/tex] is [tex]1-2(0.0384)=0.9232[/tex] which is also the probability of lifespan between 1440 and 1465


Ver imagen merlynthewhizz