Answer:43.29 hr
Explanation:
Given
mean [tex]\mu =10 hr[/tex]
standard deviation [tex]\sigma =1 hr[/tex]
n=4
Total Life time of four batteries [tex]=4\cdot \mu =4\cdot 10=40 hr[/tex]
[tex]P\left [\frac{x-\mu }{\frac{\sigma }{\sqrt{n}}}> \frac{x-40}{\frac{1}{\sqrt{4}}}\right ]=5%[/tex]
[tex]P\left [ z> \frac{x-40}{\frac{1}{\sqrt{4}}}\right ]=0.05[/tex]
[tex]1-P\left [ z< \frac{x-40}{\frac{1}{\sqrt{4}}}\right ]=0.05[/tex]
[tex]P\left [ z< \frac{x-40}{\frac{1}{\sqrt{4}}}\right ]=0.95[/tex]
from z table limiting value of [tex]z=1.645[/tex]
thus [tex]\frac{x-40}{2}=1.645[/tex]
[tex]x-40=3.29[/tex]
[tex]x=43.29[/tex]
