Question

A machine at katz steel corporation makes 3-inch-long nails. the probability distribution of the lengths of these nails is normal with a mean of 3 inches and a standard deviation of 0.1 inch. the quality control inspector takes a sample of 25 nails once a week and calculates the mean length of these nails. if the mean of this sample is either less than 2.95 inches or greater than 3.05 inches, the inspector concludes that the machine needs an adjustment.

What is the probability that based on a sample of 25 nails, the inspector will conclude that the machine needs an adjustment?

Answer 1

The probability that the inspector will conclude that the machine needs an adjustment based on a sample of 25 nails is approximately 0.617 or 61.7%.

Step-by-step explanation:

In order to find the probability that the inspector will conclude that the machine needs an adjustment based on a sample of 25 nails, we need to find the probability that the mean length of the sample is either less than 2.95 inches or greater than 3.05 inches.

We can use the Central Limit Theorem to approximate the distribution of the sample mean. The theorem states that for a large enough sample size, the distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution.

In this case, the sample size of 25 nails is large enough to use the Central Limit Theorem. The mean of the sample is the same as the mean of the population, which is 3 inches. The standard deviation of the sample mean can be calculated by dividing the standard deviation of the population by the square root of the sample size.

The standard deviation of the population is 0.1 inch, so the standard deviation of the sample mean is 0.1 / sqrt(25) = 0.02 inch.

Next, we can calculate the z-scores for the lower and upper limits of the desired range. The z-score is calculated by subtracting the mean of the sample from the desired value and dividing by the standard deviation of the sample mean.

For the lower limit of 2.95 inches, the z-score is (2.95 - 3) / 0.02 = -0.5. Using a standard normal distribution table, we find that the probability of a z-score less than -0.5 is approximately 0.3085.

For the upper limit of 3.05 inches, the z-score is (3.05 - 3) / 0.02 = 0.5. The probability of a z-score greater than 0.5 is also approximately 0.3085.

Since the events of the mean length being less than 2.95 inches and greater than 3.05 inches are mutually exclusive, we can add the probabilities together to find the probability that the inspector will conclude that the machine needs an adjustment: 0.3085 + 0.3085 = 0.617.

Therefore, the probability that based on a sample of 25 nails, the inspector will conclude that the machine needs an adjustment is approximately 0.617 or 61.7%.