Answer:
0.4364 = 43.64% probability that the disposal of such capacitors will be regulated.
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 normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean of 48.3 ppm and a standard deviation of 8 ppm.
This means that ![\mu = 48.3, \sigma = 8](https://tex.z-dn.net/?f=%5Cmu%20%3D%2048.3%2C%20%5Csigma%20%3D%208)
Sample of 40:
This means that ![n = 40, s = \frac{8}{\sqrt{40}}](https://tex.z-dn.net/?f=n%20%3D%2040%2C%20s%20%3D%20%5Cfrac%7B8%7D%7B%5Csqrt%7B40%7D%7D)
Find the probability that the disposal of such capacitors will be regulated.
Sample mean above 48.5, which is 1 subtracted by the p-value of Z when X = 48.5. 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{48.5 - 48.3}{\frac{8}{\sqrt{40}}}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B48.5%20-%2048.3%7D%7B%5Cfrac%7B8%7D%7B%5Csqrt%7B40%7D%7D%7D)
![Z = 0.16](https://tex.z-dn.net/?f=Z%20%3D%200.16)
has a p-value of 0.5636.
1 - 0.5636 = 0.4364
0.4364 = 43.64% probability that the disposal of such capacitors will be regulated.