Question

In 2017, the entire fleet of light‑duty vehicles sold in the United States by each manufacturer must emit an average of no more than 84 milligrams per mile (mg/mi) of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) over the useful life ( 150,000 miles of driving) of the vehicle. NOX + NMOG emissions over the useful life for one car model vary Normally with mean 78 mg/mi and standard deviation 6 mg/mi.

(a) What is the probability that a single car of this model emits more than 84 mg/mi of NOX + NMOG? (Enter your answer rounded to four decimal places.)

(b) A company has 3636 cars of this model in its fleet. What is the probability that the average NOX ++ NMOG level ?¯x¯ of these cars is above 8484 mg/mi? (Enter your answer rounded to four decimal places.)

Answer 1

Solution :

Given that ,

mean = $\mu$ = 78

standard deviation = $\sigma$ = 6

P(x >84 ) = 1 - P(x < 84)

= 1 - P[(x - $\mu$ ) / $\sigma$ < (84 - 78) / 6]

= 1 - P(z <1 )

Using z table,

= 1 -0.8413

=0.1587

(B)n = 36

$\mu$_{$\bar x$} = 78

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n =6 / $\sqrt$ 36= 1

P($\bar x$ >84 ) = 1 - P($\bar x$ <84 )

= 1 - P[($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (84 - 78) / 1]

= 1 - P(z <6 )

Using z table,    

= 1 - 1

= 0