Answer:
If the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 15
Standard Deviation, σ = 1
Sample size = 4
Total lifetime of 4 batteries = 40 hours
We are given that the distribution of lifetime is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Standard error due to sampling:
[tex]\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt4} = 0.5[/tex]
We have to find the value of x such that the probability is 0.05
P(X > x) = 0.05
[tex]P( X > x) = P( z > \displaystyle\frac{x - 40}{0.5})=0.05[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 40}{0.5})=0.05 [/tex]
[tex]=P( z \leq \displaystyle\frac{x - 40}{0.5})=0.95 [/tex]
Calculation the value from standard normal z table, we have,
[tex]\displaystyle\frac{x - 40}{0.5} = 1.64\\x = 40.825 \approx 40.83[/tex]
Hence, if the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.
