Question

Depths of pits on a corroded steel surface are normally distributed with mean 818 μm and standard deviation 29 μm.

A) Find the 10th percentile of pit depths.
B) A certain pit is 780 μm deep. What percentile is it on? (Round up the final answer to the nearest whole number.)
C) What proportion of pits have depths between 800 and 830 μm?

Answer 1

The 10th percentile of pit depths is approximately 780 μm. A pit depth of 780 μm is on the 9th percentile. The proportion of pits with depths between 800 and 830 μm is approximately 40.14%.

Step-by-step explanation:

To find the 10th percentile of pit depths, we can use the z-score formula. First, we calculate the z-score using the formula z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. In this case, x = 818, μ = 818, and σ = 29. Plugging these values in, we get z = (818 - 818) / 29 = 0. To find the corresponding percentile, we can look up the z-score in a z-table. The 10th percentile corresponds to a z-score of approximately -1.2816. To find the value, we can use the formula z = (x - μ) / σ and solve for x. Rearranging the formula, we have x = μ + zσ = 818 + (-1.2816) * 29 = 780.35 ≈ 780. Therefore, the 10th percentile of pit depths is approximately 780 μm.

To find the percentile that a certain pit depth of 780 μm is on, we can use the z-score formula and the same steps as before. Plugging in the values, we get z = (780 - 818) / 29 = -1.3103. Looking up this z-score in the z-table, we find that it corresponds to a percentile of approximately 9%. Therefore, the pit depth of 780 μm is on the 9th percentile.

To find the proportion of pits with depths between 800 and 830 μm, we can first find the z-scores for these values using the same formula. Plugging in the values, we get z1 = (800 - 818) / 29 = -0.6207 and z2 = (830 - 818) / 29 = 0.4138. To find the proportion between these z-scores, we can subtract the cumulative area corresponding to z1 from the cumulative area corresponding to z2. Looking up these z-scores in the z-table, we find that the cumulative areas are approximately 26.41% and 66.55%, respectively. Therefore, the proportion of pits with depths between 800 and 830 μm is approximately 66.55% - 26.41% = 40.14%.

Answer 2

The student's questions involve using the properties of the normal distribution to determine percentiles and the proportion of values within a given range for pit depths on a corroded steel surface. This involves calculating Z-scores and using the Z-table to find corresponding percentiles and probabilities.

Step-by-step explanation:

The student is confronted with a series of problems dealing with the normal distribution and its properties, specifically in the context of pit depths in a corroded steel surface.

A) Find the 10th percentile of pit depths

To find the 10th percentile, we use the Z-table. First, find the Z-score that corresponds to the 10th percentile. Then use the formula Z = (X - μ) / σ, where X is the pit depth, μ is the mean, and σ is the standard deviation. Plugging in the values μ = 818 and σ = 29, solve for X.

B) What percentile is a pit that is 780 μm deep?

First calculate the Z-score for a pit depth of 780 μm. Then find the corresponding percentile using the Z-table. This percentile is the rank of the pit depth among all possible depths.

C) What proportion of pits have depths between 800 and 830 μm?

To find the proportion of pits within a specific range, calculate the Z-scores for both 800 and 830 μm. Look up these Z-scores in the Z-table to find the cumulative probabilities. The proportion of pits in this range is the difference between these two probabilities.