Answer:
a) 0.2119 = 21.19% probability that a single car of this model emits more than 86 mg/mi of NOX +NMOG
b) 0% probability that the average NOX NMOG level cars is above 86 mg/mi
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
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 normally distributed random variable X, with mean and standard deviation , the sample means with size n 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 emits more than 86 mg/mi of NOX +NMOG?
This is 1 subtracted by the pvalue of Z when X = 86. So
has a pvalue of 0.7881
1 - 0.7881 = 0.2119
0.2119 = 21.19% probability that a single car of this model emits more than 86 mg/mi of NOX +NMOG
(b) A company has 25 cars of this model in its fleet. What is the probability that the average NOX NMOG level cars is above 86 mg/mi?
Now we have that
This is 1 subtracted by the pvalue of Z when X = 86. So
By the Central Limit Theorem
has a pvalue of 0.99998.
1-99998 = 0.00002
0% probability that the average NOX NMOG level cars is above 86 mg/mi