Answer:
The probability that a light bulb of that brand lasts between 1175 hr and 1610 hr is 0.8524.
Step-by-step explanation:
Given : Suppose a brand of light bulbs is normally distributed, with a mean life of 1400 hr and a standard deviation of 150 hr.
To find : The probability that a light bulb of that brand lasts between 1175 hr and 1610 hr ?
Solution :
Applying z-score formula,
[tex]z=\frac{x-\mu}{\sigma}[/tex]
where, [tex]\mu=1400[/tex] is population mean
[tex]\sigma=150[/tex] is standard deviation
For x=1175 hour,
[tex]z=\frac{1175-1400}{150}[/tex]
[tex]z=\frac{-225}{150}[/tex]
[tex]z=-1.5[/tex]
For x=1610 hour,
[tex]z=\frac{1610-1400}{150}[/tex]
[tex]z=\frac{210}{150}[/tex]
[tex]z=1.4[/tex]
The required probability is,
[tex]P(1175< X<1610)=P(-1.5<z<1.4)[/tex]
[tex]P(1175< X<1610)=P(z<1.4)-P(z<-1.5)[/tex]
Using z table, the values are
[tex]P(1175< X<1610)=0.9192-0.0668[/tex]
[tex]P(1175< X<1610)=0.8524[/tex]
The probability that a light bulb of that brand lasts between 1175 hr and 1610 hr is 0.8524.
