Question

A battery company claims that their batteries last on average 60 hours. The standard deviation of the battery life is 4 hours. Assuming that their claim is true, what would be the lowest 5% of values we would get if we were to sample 100 batteries and find their mean life length?

Answer 1

Given that,

mean = = 60

standard deviation = = 4

n = 100

= 60

= / n =4 /100=0.4

Using standard normal table,

P(Z < z) = 5%

= P(Z < z) = 0.05

= P(Z < -1.64) = 0.05

z =-1.64 Using standard normal table,

Using z-score formula  

= z * +   

= -1.64 *0.4+60

= 59.344