Mathematics: 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.)

The level of nitrogen oxides (nox) in the exhaust

Ответ:

ewilly01

15.03.2022

0.533

Explanation: Note that

$x \ \textgreater \ I \\ \Leftrightarrow x - 0.3 \ \textgreater \ I - 0.3 \\ \\ \Leftrightarrow \boxed{\frac{x - 0.3}{0.1} \ \textgreater \ \frac{I - 0.3}{0.1}} (1)$

Let

z_1

= z-score for I.
z_2

= z-score for x.

The formula for z-score is given by

$\text{z-score} = \frac{(\text{data point}) - (\text{mean})}{(\text{standard deviation})}$

So,

$z_2 = \frac{x - 0.3}{0.1}$

$z_1 = \frac{I - 0.3}{0.1}$

Based on inequality in (1),

$z_2 \ z_1$

So,

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

Since z_1

and

are z-scores and the level of nitrogen oxides are normally distributed, using normal distribution calculator (or table),

$z_1 = 2.326
\\
\\ \Leftrightarrow \frac{I - 0.3}{0.1} = 2.326
\\
\\ \Leftrightarrow I - 0.3 = 0.1(2.326)
\\
\\ \Leftrightarrow I - 0.03 = 0.2326
\\
\\ \Leftrightarrow \boxed{I \approx 0.533}$

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Ответ:

malika61

10.06.2020

Mathematics

5 hours

Step-by-step explanation:

If 10% of 140 is 14. And if 14 mg of caffeine is leaving per hour then it would take 10 hours to completely leave from ones body, but for half of it to leave it would take 5.

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