Question

Polychlorinated biphenyl (PCB) is an organic pollutant that can be found in electrical equipment. A certain kind of small capacitor contains PCB with a mean of 48.2 ppm (parts per million) and a standard deviation of 8 ppm. A governmental agency takes a random sample of 39 of these small a capacitors. The agency plans to regulate the disposal of such capacitors if the sample mean amount of PCB is 49.5 ppm or more. Find the probability that the disposal of such capacitors will be regulated Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.

Answer 1

To find the probability that the disposal of such capacitors will be regulated, we need to calculate the probability of obtaining a sample mean of 49.5 ppm or more. Using the z-score formula and a standard normal distribution table or calculator, we find the probability is approximately 0.0981.

Step-by-step explanation:

To find the probability that the disposal of such capacitors will be regulated, we need to calculate the probability of obtaining a sample mean of 49.5 ppm or more. We can use the z-score formula to do this. The z-score is calculated as: (sample mean - population mean) / (standard deviation / sqrt(sample size)). Plugging in the given values:

z = (49.5 - 48.2) / (8 / sqrt(39))

z ≈ 1.286

Next, we need to find the probability corresponding to this z-score. We can use a standard normal distribution table or a calculator to find this probability. In this case, the probability can be calculated as the area under the standard normal curve to the right of the z-score of 1.286. Using a standard normal distribution table or calculator, we find that the probability is approximately 0.0981, rounded to four decimal places.

Answer 2

The probability that the disposal of capacitors with a sample mean of PCB concentration of 49.5 ppm or higher will be regulated is approximately 0.845, calculated using the z-score and the standard normal distribution.

Step-by-step explanation:

The student has asked about finding the probability that a random sample of 39 capacitors that contain PCBs will have a mean PCB concentration of 49.5 ppm or higher. Given that the population mean is 48.2 ppm and the standard deviation is 8 ppm, we can use the Central Limit Theorem to find this probability since the sample size is fairly large (n = 39).

First, we determine the standard error of the mean (SEM) which is the standard deviation divided by the square root of the sample size:

SEM = σ / √n = 8 ppm / √39 = 8 ppm / 6.244 = 1.2816 ppm (rounded to four decimal places)

Next, we find the z-score, which is the number of standard errors the sample mean is above the population mean. The z-score formula is:

Z = (X - μ) / SEM
Z = (49.5 - 48.2) / 1.2816 = 1.0169 (rounded to four decimal places)

We then use the standard normal distribution table or a calculator with a cumulative distribution function to find the probability that Z is greater than 1.0169. This probability corresponds to the area to the right of the z-score on the normal curve, which gives us the likelihood of regulating the disposal of such capacitors.

The exact probability can be obtained from tables or software, but for illustration, a z-score of 1.0169 typically corresponds to a probability of approximately 0.1554 to the left. To find the probability to the right (the area more than this z-score), we subtract this value from 1, resulting in:

Probability = 1 - 0.1554 = 0.8446 (rounded to four decimal places)

Therefore, the probability that the disposal of capacitors will be regulated is 0.845 (rounded to three decimal places).