Respuesta :
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:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(score greater than 0.3)
P(x > 0.3)
[tex]P( x > 0.3) = P( z > \displaystyle\frac{0.3 - 0.25}{0.05}) = P(z > 1)[/tex]
[tex]= 1 - P(z \leq 1)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 0.3) = 1 - 0.8413 = 0.1587 = 15.87\%[/tex]
0.1587 is the probability that the average nitrogen oxides level of cars is above the 0.3 g/mi limit.
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([tex]\mu=0.25,\sigma^{2}=0.05^{2}[/tex])
Now, the z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean
[tex]\sigma[/tex] = 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( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{0.30-0.25}{0.05}[/tex] ) = P(Z > 1) = 1 - P(Z [tex]\leq[/tex] 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.
