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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 ![s = \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In this problem, we have that:
![\mu = 82, \sigma = 5](https://tex.z-dn.net/?f=%5Cmu%20%3D%2082%2C%20%5Csigma%20%3D%205)
(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
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{86 - 82}{5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B86%20-%2082%7D%7B5%7D)
![Z = 0.8](https://tex.z-dn.net/?f=Z%20%3D%200.8)
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 ![n = 25, s = \frac{5}{\sqrt{25}} = 1](https://tex.z-dn.net/?f=n%20%3D%2025%2C%20s%20%3D%20%5Cfrac%7B5%7D%7B%5Csqrt%7B25%7D%7D%20%3D%201)
This is 1 subtracted by the pvalue of Z when X = 86. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{86 - 82}{1}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B86%20-%2082%7D%7B1%7D)
![Z = 4](https://tex.z-dn.net/?f=Z%20%3D%204)
has a pvalue of 0.99998.
1-99998 = 0.00002
0% probability that the average NOX NMOG level cars is above 86 mg/mi