The level of nitrogen oxides (NO x ) in the exhaust of cars of a particular model varies Normally, with mean 0.25 gram per mile (g/mi) and s

Question

The level of nitrogen oxides (NO x ) in the exhaust of cars of a particular model varies Normally, with mean 0.25 gram per mile (g/mi) and standard deviation 0.05 g/mi. Government regulations call for NO x emissions no greater than 0.3 g/mi. A company has four cars of this model in its fleet. What is the probability that the average NO x level of these cars is above the 0.3 g/mi limit?

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Maria 5 days 2021-10-08T11:06:36+00:00 2 Answers 0

Answers ( )

    0
    2021-10-08T11:07:53+00:00

    Answer:

    0.1587 is the probability that the average nitrogen oxides level of cars is above the 0.3 g/mi limit.

    Step-by-step explanation:

    We are given the following information in the question:

    Mean, μ = 0.25 gram per mile

    Standard Deviation, σ = 0.05 g/m

    We are given that the distribution of level of nitrogen oxides is a bell shaped distribution that is a normal distribution.

    Formula:

    z_{score} = \displaystyle\frac{x-\mu}{\sigma}

    P(score greater than 0.3)

    P(x > 0.3)

    P( x > 0.3) = P( z > \displaystyle\frac{0.3 - 0.25}{0.05}) = P(z > 1)

    = 1 - P(z \leq 1)

    Calculation the value from standard normal z table, we have,  

    P(x > 0.3) = 1 - 0.8413 = 0.1587 = 15.87\%

    0.1587 is the probability that the average nitrogen oxides level of cars is above the 0.3 g/mi limit.

    0
    2021-10-08T11:08:09+00:00

    Answer:

    Probability that the average NO x level of these cars is above the 0.3 g/mi limit is 0.15866.

    Step-by-step explanation:

    We are given that the level of nitrogen oxides (NO x ) in the exhaust of cars of a particular model varies Normally, with mean 0.25 gram per mile (g/mi) and standard deviation 0.05 g/mi.

    Let X = level of nitrogen oxides (NO x ) in the exhaust of cars

    So, X ~ N(\mu=0.25,\sigma^{2}=0.05^{2})

    Now, the z score probability distribution is given by;

             Z = \frac{X-\mu}{\sigma} ~ N(0,1)

    where, \mu = population mean

                \sigma = standard deviation

    So, probability that the average NO x level of these cars is above the 0.3 g/mi limit is given by = P(X > 0.30 g/mi)

         P(X > 0.30) = P( \frac{X-\mu}{\sigma} > \frac{0.30-0.25}{0.05} ) = P(Z > 1) = 1 – P(Z \leq 1)

                                                              = 1 – 0.84134 = 0.15866

    Therefore, probability that the average NO x level of these cars is above the 0.3 g/mi limit is 0.15866.

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