The lifetimes of batteries produced by the juicy battery company are normally distributed with a mean of 100 hours and a standard deviation of 10 hours. find the number of hours where at least 90% of the batteries will continue to operate at.

The z-score associated with 90% is:
P(0.9)=1.28
sup[pose the number of hours whereby at least 90% of the batteries will continue to operate is x.
the z-score is given by:
z=(x-μ)/σ
where:
μ is the mean
σ is the standard deviation
substituting the values we shall have:
1.28=(x-100)/10
solving for x we get:
x=112.8 hours

 

The answer of the above is

He will need 2,898 pieces of paper