The level of nitrogen oxides (nox) in the exhaust after 50,000 miles or fewer of driving of cars of a particular model varies normally with mean 0.03 g/mi and standard deviation 0.01 g/mi. a company has 64 cars of this model in its fleet. what is the level l such that the probability that the average nox level x for the fleet is greater than l is only 0.01? (hint: this requires a backward normal calculation. round your answer to three decimal places.)

Answer: 0.533

Explanation: Note that

[tex]x \ \textgreater \ I \\ \Leftrightarrow x - 0.3 \ \textgreater \ I - 0.3 \\ \\ \Leftrightarrow \boxed{\frac{x - 0.3}{0.1} \ \textgreater \ \frac{I - 0.3}{0.1}} (1) [/tex] 

Let

[tex]z_1[/tex] = z-score for I.
[tex]z_2[/tex] = z-score for x. 

The formula for z-score is given by

[tex]\text{z-score} = \frac{(\text{data point}) - (\text{mean})}{(\text{standard deviation})} [/tex]

So,

[tex]z_2 = \frac{x - 0.3}{0.1}[/tex]

[tex]z_1 = \frac{I - 0.3}{0.1}[/tex]

Based on inequality in (1), 

[tex]z_2 > \ z_1[/tex] 

So, 

[tex]P(x \ \textgreater \ I) = 0.01 \\ \\ \Leftrightarrow P \left (\frac{x - 0.3}{0.1} \ \textgreater \ \frac{I - 0.3}{0.1} \right ) = 0.01 \\ \\ \Leftrightarrow P(z_2 \ \textgreater \ z_1) = 0.01 \\ \\ \Leftrightarrow 1 - P(z_2 \leq z_1) = 0.01 \\ \\ \Leftrightarrow \boxed{P(z_2 \leq z_1) = 0.99}[/tex]

Since [tex]z_1[/tex] and [tex]z_2[/tex] are z-scores and the level of nitrogen oxides are normally distributed, using normal distribution calculator (or table), 

[tex]z_1 = 2.326 \\ \\ \Leftrightarrow \frac{I - 0.3}{0.1} = 2.326 \\ \\ \Leftrightarrow I - 0.3 = 0.1(2.326) \\ \\ \Leftrightarrow I - 0.03 = 0.2326 \\ \\ \Leftrightarrow \boxed{I \approx 0.533}[/tex]