Question

The level of nitrous oxide in the exhaust after 50000 miles in a car of a certain model follows Normal distribution with mean 0.03g/mi and standard deviation 0.01g/mi. A company has 25 cars of this model in its fleet. What is the level L such that the probability that the average nitrous oxide level for the fleet is greater than L is 0.01

Answer 1

The value is
`L = 0.035 \ g/mi`

Explanation:

From the question we are told that

The mean is
`\mu = 0.03 \ g/mi`

The standard deviation is
`\sigma = 0.01 \ g/mi`

The sample size is n = 25

Generally the standard error of mean is mathematically represented as

`\sigma_(x) = (\sigma )/(√(n) )`

=>
`\sigma_(x) = ( 0.01 )/(√(25) )`

=>
`\sigma_(x) =0.002`

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

`P(X > L ) = P(( X - \mu )/(\sigma_(x)) > (L- 0.03)/(0.002 ) ) =0.01`

`(X -\mu)/(\sigma )  =  Z (The  \ standardized \  value\  of  \ X )`

`P(X > L ) = P(Z > (L- 0.03)/(0.002 ) ) =0.01`

Note

`P(X > L ) = P(Z > (L- 0.03)/(0.002 ) ) =0.01 \equiv P(Z < z ) = 0.01`

From the normal distribution table the critical value of 0.01 is

`z = 2.326`

Hence

`(L- 0.03)/(0.002 ) = 2.326`

=>
`L = 0.035 \ g/mi`