Respuesta :
Answer:
a) 15.87% probability that a single car of this model fails to meet the NOX requirement.
b) 2.28% probability that the average NOX level of these cars are above 0.3 g/mi limit
Step-by-step explanation:
We use the normal probability distribution and the central limit theorem to solve this question.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 0.25, \sigma = 0.05[/tex]
a. What is the probability that a single car of this model fails to meet the NOX requirement?
Emissions higher than 0.3, which is 1 subtracted by the pvalue of Z when X = 0.3. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.3 - 0.25}{0.05}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8417.
1 - 0.8413 = 0.1587.
15.87% probability that a single car of this model fails to meet the NOX requirement.
b. A company has 4 cars of this model in its fleet. What is the probability that the average NOX level of these cars are above 0.3 g/mi limit?
Now we have [tex]n = 4, s = \frac{0.05}{\sqrt{4}} = 0.025[/tex]
The probability is 1 subtracted by the pvalue of Z when X = 0.3. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.3 - 0.25}{0.025}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
1 - 0.9772 = 0.0228
2.28% probability that the average NOX level of these cars are above 0.3 g/mi limit
The probability that a single car of this model fails to meet the NOX requirement is 15.87%.
What is z score?
Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
z = (raw score - mean) / standard deviation
Given; mean of 0.25 g and a standard deviation of 0.05 g/mi
a) For > 0.3:
z = (0.3 - 0.25)/0.05 = 1
P(z > 1) = 1 - P(z < 1) = 1 - 0.8413 = 0.1587
b) For > 0.3, sample size = 4
z = (0.3 - 0.25)/(0.05 ÷√4) = 2
P(z > 2) = 1 - P(z < 2) = 1 - 0.9772 = 0.0228
The probability that a single car of this model fails to meet the NOX requirement is 15.87%.
Find out more on z score at: https://brainly.com/question/25638875
