Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
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 [tex]\mu = 48.3, \sigma = 8[/tex]
Sample of 40:
This means that [tex]n = 40, s = \frac{8}{\sqrt{40}}[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{48.5 - 48.3}{\frac{8}{\sqrt{40}}}[/tex]
[tex]Z = 0.16[/tex]
[tex]Z = 0.16[/tex] 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.
