Question

Polychlorinated biphenyl (PCB) is among a group of organic pollutants found in a variety of products, such as coolants, insulating materials, and lubricants in electrical equipment. Disposal of items containing less than 50 parts per million (ppm) PCB is generally not regulated. A certain kind of small capacitor contains PCB with a mean of 48.3 ppm and a standard deviation of 8 ppm. The Environmental Protection Agency takes a random sample of 40 of these small capacitors, planning to regulate the disposal of such capacitors if the sample mean amount of PCB is 48.5 ppm or more. Find the probability that the disposal of such capacitors will be regulated. Carry your intermediate comp

Answer 1

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 $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

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 $\mu = 48.3, \sigma = 8$

Sample of 40:

This means that $n = 40, s = \frac{8}{\sqrt{40}}$

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

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{48.5 - 48.3}{\frac{8}{\sqrt{40}}}$

$Z = 0.16$

$Z = 0.16$ 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.