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 47.5 ppm (parts per million) and a standard deviation of 6 ppm. A governmental agency takes a random sample of 40 of these small capacitors. The agency plans 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 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 that the sample mean amount of pcb is 48.5 ppm or more. The probability is approximately 0.1471 or 14.71%.

Step-by-step explanation:

To find the probability that the disposal of such capacitors will be regulated, we need to calculate the probability that the sample mean amount of pcb is 48.5 ppm or more.

First, we need to calculate the standard error of the sample mean. The standard error can be calculated by dividing the standard deviation by the square root of the sample size.

Standard error = standard deviation / √sample size

In this case, the standard error = 6 ppm / √40 = 0.9487 ppm.

Next, we need to calculate the z-score, which represents the number of standard errors away from the mean the sample mean falls.

Z-score = (sample mean - population mean) / standard error

In this case, the z-score = (48.5 ppm - 47.5 ppm) / 0.9487 ppm = 1.0542

Finally, we can use a standard normal distribution table or a calculator to find the probability that the z-score is greater than or equal to 1.0542. From the table or calculator, we find that the probability is approximately 0.1471.

Therefore, the probability that the disposal of such capacitors will be regulated is approximately 0.1471 or 14.71%.