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?

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:

P(score greater than 0.3)

P(x > 0.3)

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

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

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()

Now, the z score probability distribution is given by;

         Z = ~ N(0,1)

where, = population mean

            = 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( > ) = P(Z > 1) = 1 - P(Z 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.

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation: