Problem 195 - Applied Statistics and Probability for Engineers

Question

The length of an injection-molded plastic case that holds magnetic tape is normally distributed with a length of 90.2 millimeters and a standard deviation of 0.1 millimeter. (a) What is the probability that a part is longer than 90.3 millimeters or shorter than 89.7 millimeters? (b) What should the process mean be set at to obtain the highest number of parts between 89.7 and 90.3 millimeters? (c) If parts that are not between 89.7 and 90.3 millimeters are scrapped, what is the yield for the process mean that you selected in part (b)? Assume that the process is centered so that the mean is 90 millimeters and the standard deviation is 0.1 millimeter. Suppose that 10 cases are measured, and they are assumed to be independent. (d) What is the probability that all 10 cases are between 89.7 and 90.3 millimeters? (e) What is the expected number of the 10 cases that are between 89.7 and 90.3 millimeters?

Step-by-Step Solution

Answer

(a) ~0.1587; (b) 90.0 mm; (c) 84.13% yield; (d) ~0.193; (e) ~8.4 cases.

Step 1: Calculate Z-Scores for Given Lengths

We need to find the Z-scores for 90.3 and 89.7 millimeters to determine the probability that a part is longer than 90.3 millimeters or shorter than 89.7 millimeters. The Z-score formula is \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the length, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.- For 90.3 mm: \( Z = \frac{90.3 - 90.2}{0.1} = 1.0 \).- For 89.7 mm: \( Z = \frac{89.7 - 90.2}{0.1} = -5.0 \).

Step 2: Find Probabilities for Calculated Z-Scores

Using the standard normal distribution table, find the probabilities corresponding to the calculated Z-scores:- The probability for \( Z > 1.0 \) is roughly 0.1587 (since the table gives \( P(Z < 1.0) = 0.8413 \)).- The probability for \( Z < -5.0 \) is essentially 0, as values beyond the Z-table are negligible.Hence, the probability that a part is longer than 90.3 mm or shorter than 89.7 mm is approximately \( 0.1587 \).

Step 3: Optimal Process Mean for Maximum Yield

To maximize the yield of parts between 89.7 mm and 90.3 mm, the process mean should be set at the midpoint of these two values, which naturally centers the distribution.The midpoint is \( \frac{89.7 + 90.3}{2} = 90.0 \) mm. Thus, the mean should be set to 90.0 mm.

Step 4: Calculate Yield with Mean at 90 mm

With the mean set at 90 mm (as chosen in Step 3), calculate the yield:The probabilities for lengths between 89.7 mm and 90.3 mm using the Z-scores from Step 1 are as follows:- \( P(-5.0 < Z < 1.0) \) corresponds to the normal distribution range between these Z-scores, which yields \( P(-5.0 < Z < 1.0) = 0 + 0.8413 = 0.8413 \).Thus, the yield is 84.13%.

Step 5: Probability of All 10 Cases Within Limits

Given that the probability of one case being between 89.7 mm and 90.3 mm is 0.8413, the probability that all 10 cases meet this criteria is given by:\[ P(\text{all 10 cases}) = (0.8413)^{10} \approx 0.193 \, . \]

Step 6: Expected Number of Cases Within Limits

The expected number of cases within 89.7 mm to 90.3 mm can be calculated using the formula \( E = n \cdot p \), where \( n \) is the number of trials (10 cases), and \( p \) is the probability of a single event (0.8413):\[ E = 10 \cdot 0.8413 = 8.413 \].

Key Concepts

Normal Distribution

The normal distribution, often referred to as the bell curve, is a fundamental concept in statistics. It's a probability distribution that is symmetric around its mean. Real-world phenomena, such as human heights or test scores, often follow this distribution. The properties of the normal distribution are:

• The mean, median, and mode are all equal.
• The curve is symmetric about the mean.
• It follows the empirical rule, which means about 68% of values lie within one standard deviation of the mean, 95% within two, and 99.7% within three.

Understanding this can help when assessing probability problems, like determining the likelihood of a part being within a specific measurement range.

Z-Score

A Z-score tells you how many standard deviations a particular data point is from the mean of a data set. It is a crucial concept when dealing with normal distributions because it helps standardize different data points, allowing for comparison across different distributions.To calculate a Z-score, use the formula:\[ Z = \frac{X - \mu}{\sigma} \] where:

• \( X \) is the value of the data point,
• \( \mu \) is the mean of the data set, and
• \( \sigma \) is the standard deviation.

A positive Z-score means the data point is above the mean, while a negative one indicates it's below the mean. This concept is particularly useful for determining probabilities and identifying outliers in data.

Process Yield

Process yield refers to the percentage of produced units that meet quality criteria without needing reworks. In manufacturing or quality control, a high process yield indicates that most units produced are within the desired specifications. In the context of normally distributed data, the yield can be computed by determining the proportion of outcomes within a specific range. Understanding process yield is critical for efficiency and profitability, as low yield could indicate issues in the production process that need addressing to reduce waste and increase output.

Expected Value

Expected value is a concept used to determine the average outcome if a situation was repeated multiple times. It is especially useful in probabilistic assessments and risk analysis. To calculate the expected value in a discrete probability distribution, use the formula:\[ E(X) = \sum P(x) \cdot x \] where:

• \( E(X) \) is the expected value,
• \( P(x) \) is the probability of each outcome, and
• \( x \) is the value of each outcome.

In problems involving measurements, such as our example with part lengths, it helps to predict the average number of units falling within a specified range, aiding in planning and expectation setting.