Answer:
The value is  [tex]L = 0.035 \ g/mi[/tex]
Step-by-step explanation:
From the question we are told that
  The mean is [tex]\mu = 0.03 \ g/mi[/tex]
  The standard deviation is  [tex]\sigma = 0.01 \ g/mi[/tex]
  The sample size is  n  = 25
Generally the standard error of mean is mathematically represented as
   [tex]\sigma_{x} = \frac{\sigma }{\sqrt{n} }[/tex]
=> Â [tex]\sigma_{x} = \frac{ 0.01 }{\sqrt{25} }[/tex] Â
=> Â [tex]\sigma_{x} =0.002[/tex]
Generally the  the level L such that the probability that the average nitrous oxide level for the fleet is greater than L is 0.01 is mathematically evaluated as
  [tex]P(X > L ) = P(\frac{ X - \mu }{\sigma_{x}} > \frac{L- 0.03}{0.002 } ) =0.01[/tex]
[tex]\frac{X -\mu}{\sigma }  =  Z (The  \ standardized \  value\  of  \ X )[/tex]
 [tex]P(X > L ) = P(Z > \frac{L- 0.03}{0.002 } ) =0.01[/tex]
Note
[tex]P(X > L ) = P(Z > \frac{L- 0.03}{0.002 } ) =0.01 \equiv P(Z < z ) = 0.01[/tex]
From the normal distribution table the critical value of  0.01  is Â
   [tex]z = 2.326[/tex]
Hence Â
    [tex]\frac{L- 0.03}{0.002 } = 2.326[/tex]
=> Â [tex]L = 0.035 \ g/mi[/tex]
