Answer: 0.0478
Step-by-step explanation:
Given : The life of a particular brand of laptop battery is normally distributed with a mean of 8 hours and a standard deviation of 0.6 hours.
i.e. [tex]\mu=8[/tex]
[tex]\sigma=0.6[/tex]
Let x be the random variable that represents the life of a particular brand of laptop battery.
Then , the probability that the battery will last more than 9 hours before running out of power will be :-
[tex]P(x>9)=P(z>\dfrac{9-8}{0.6})[/tex] [∵ [tex]z=\dfrac{x-\mu}{\sigma}[/tex]]
[tex]=P(z>1.6667)=1-P(z\leq1.6667)[/tex] [∵ P(Z>z)=1-P(Z≤z)]
[tex]=1-0.9522=0.0478[/tex] [using P-value table for z]
Hence, the probability that the battery will last more than 9 hours before running out of power = 0.0478
