I need help with 89 divided by 6 in remainder
2 answers:
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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 and standard deviation
, the zscore of a measure X is given by:
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 and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
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
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
The probability is 1 subtracted by the pvalue of Z when X = 0.3. So
By the Central Limit Theorem
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