Problem 211 - Applied Statistics and Probability for Engineers

Question

The thickness of a laminated covering for a wood surface is normally distributed with a mean of five millimeters and a standard deviation of 0.2 millimeter. (a) What is the probability that a covering thickness is more than 5.5 millimeters? (b) If the specifications require the thickness to be between 4.5 and 5.5 millimeters, what proportion of coverings does not meet specifications? (c) The covering thickness of \(95 \%\) of samples is below what value?

Step-by-Step Solution

Answer

(a) 0.0062; (b) 0.0124; (c) 5.329 mm

Step 1: Understand the Distribution

The thickness of the laminated covering is normally distributed with a mean (\(\mu\)) of 5 mm and a standard deviation (\(\sigma\)) of 0.2 mm. We will use these parameters to solve each part of the problem using properties of the normal distribution.

Step 2: Calculate Z-Score for 5.5 mm (Part a)

To find the probability that a covering thickness is more than 5.5 mm, calculate the Z-score. The Z-score formula is: \[ Z = \frac{X - \mu}{\sigma} \] Substitute the given values: \(X = 5.5\), \(\mu = 5\), \(\sigma = 0.2\): \[ Z = \frac{5.5 - 5}{0.2} = 2.5 \].

Step 3: Find Probability for Z-Score (Part a)

Using the standard normal distribution table, find the probability corresponding to the Z-score of 2.5. The table gives the probability that a value is less than 5.5 mm, which is approximately 0.9938. Therefore, the probability that it is more than 5.5 mm is \(1 - 0.9938 = 0.0062\).

Step 4: Calculate Z-Scores for 4.5 mm and 5.5 mm (Part b)

For 4.5 mm, calculate the Z-score: \[ Z = \frac{4.5 - 5}{0.2} = -2.5 \]. For 5.5 mm, the Z-score is already found to be 2.5.

Step 5: Find Proportion within 4.5 mm and 5.5 mm (Part b)

Using the normal distribution table:- For \(Z = -2.5\), the probability is approximately 0.0062.- For \(Z = 2.5\), the probability is 0.9938.The proportion within 4.5 mm and 5.5 mm is \(0.9938 - 0.0062 = 0.9876\). Therefore, the proportion that does not meet specifications is \(1 - 0.9876 = 0.0124\).

Step 6: Find Z-Score for 95% (Part c)

For the 95th percentile, find the Z-score that corresponds to 0.95 probability in the standard normal distribution table, which is approximately 1.645.

Step 7: Calculate Covering Thickness for 95% (Part c)

To find the thickness value, use the formula: \[ X = \mu + Z \times \sigma \]. Substitute \(\mu = 5\), \(Z = 1.645\), and \(\sigma = 0.2\): \[ X = 5 + 1.645 \times 0.2 = 5.329 \text{ mm} \].

Key Concepts

Probability Calculation

Probability calculation is a fundamental concept in statistics that helps us determine how likely an event is to occur. When dealing with normally distributed data, such as the thickness of the laminated covering, we can calculate probabilities using Z-scores.
In part (a) of the exercise, we are asked to find the probability that the thickness is more than 5.5 mm. First, we calculated the Z-score for 5.5 mm, which turned out to be 2.5. The Z-score tells us how many standard deviations away from the mean our value is. With this Z-score, using a standard normal distribution table, we found the probability of the thickness being less than 5.5 mm. Since the problem asks for the probability of being more than 5.5 mm, we subtracted the found probability from 1, which is a common practice when calculating probabilities for events greater than a given value. Doing these steps, we arrived at the probability of 0.0062, or 0.62%, indicating a rare event.
Understanding probability calculation helps us make informed decisions based on statistical data, such as quality control in manufacturing processes.

Z-Score

A Z-score is a measure that describes a value's position relative to the mean of a group of values. It's expressed in terms of the number of standard deviations away from the mean.
Mathematically, the Z-score is calculated as:
\[ Z = \frac{X - \mu}{\sigma} \] where

• \(X\) is the value of interest,
• \(\mu\) is the mean,
• \(\sigma\) is the standard deviation.

For example, in part (b), the exercise calculates Z-scores for thickness values of 4.5 mm and 5.5 mm. This helps to determine the probability within these specifications. The Z-scores were -2.5 and 2.5, meaning 4.5 mm is 2.5 standard deviations below the mean and 5.5 mm is 2.5 standard deviations above.
Using these Z-scores, we can refer to standard normal distribution tables to find the corresponding probabilities. Z-scores are a powerful tool for comparing different values within a normal distribution and understanding their deviations from the average.

Standard Deviation

The standard deviation is a key concept in statistics that measures the amount of variation or dispersion in a set of values. In a normal distribution, the standard deviation helps us understand how spread out the values are around the mean.
In this exercise, the standard deviation was given as 0.2 mm. This small standard deviation indicates that the thickness values are tightly clustered around the mean of 5 mm.
When calculating probabilities and Z-scores, the standard deviation is crucial because it scales the difference between the value and the mean by this spread measure. This is evident in part (c) where, to find the thickness value below which 95% of the samples lie, we used the standard deviation along with a Z-score to compute this critical value.
Ultimately, understanding standard deviation aids in assessing how much variance exists in our measurements and in making predictions about future or unseen data points.