Tensile Strength of Wires: Statistical Analysis

Explanation

95% of the observations lie between plus and minus two standard deviations from the mean, so it is 500 +/- 2(40).

To determine the range within which the middle 95 percent of the wires tested fall, we need to calculate the confidence interval. Since we have the mean and standard deviation, we can use the normal distribution to estimate the range.

The mean of the sample is given as 500 PSI, which indicates the average tensile strength of the wires. The median and mode are also reported as 500 PSI, suggesting that the data is symmetrically distributed.

The standard deviation is given as 40 PSI, which measures the spread or dispersion of the data points around the mean. A larger standard deviation implies a greater variability in the data.

To find the range that includes the middle 95 percent of the data, we can use the concept of the standard normal distribution. In a standard normal distribution, approximately 95 percent of the data falls within two standard deviations of the mean.

Since we know the mean deviation (also known as the average deviation) is given as 32 PSI, we can use it as an estimate for the standard deviation. However, note that the mean deviation is typically smaller than the standard deviation. Therefore, the actual range might be slightly wider.

To calculate the range for the middle 95 percent, we need to determine the interval that covers two standard deviations on either side of the mean. In this case, we use the mean deviation (32 PSI) as an estimate for the standard deviation.

Lower limit = Mean - (2 * Standard Deviation) Upper limit = Mean + (2 * Standard Deviation)

Lower limit = 500 - (2 * 32) = 500 - 64 = 436 PSI Upper limit = 500 + (2 * 32) = 500 + 64 = 564 PSI

Therefore, the middle 95 percent of the wires tested fall between approximately 436 PSI and 564 PSI.

Among the given answer choices, the closest range to this is option D: 460 and 540 PSI.