Problem 84 - Applied Statistics and Probability for Engineers

Question

The diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.002 inch and a standard deviation of 0.0004 inch. (a) What is the probability that the diameter of a dot exceeds \(0.0026 ?\) (b) What is the probability that a diameter is between 0.0014 and \(0.0026 ?\) (c) What standard deviation of diameters is needed so that the probability in part (b) is \(0.995 ?\)

Step-by-Step Solution

Answer

(a) 0.0668 (b) 0.8664 (c) 0.000233 inch

Step 1: Understanding the normal distribution

The diameter of dots is normally distributed with mean \( \mu = 0.002 \) inch and standard deviation \( \sigma = 0.0004 \) inch. We'll use this information for calculating different probabilities.

Step 2: Finding Z-Score for Diameter 0.0026

To find the probability that a dot's diameter exceeds 0.0026, we calculate the Z-score using the formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X = 0.0026 \). Plugging in the values: \[ Z = \frac{0.0026 - 0.002}{0.0004} = 1.5 \].

Step 3: Probability Calculation for Exceeding 0.0026

Using the Z-score from Step 2, we look up the standard normal distribution table to find the probability of Z being greater than 1.5. This probability is \(1 - P(Z < 1.5) \approx 1 - 0.9332 = 0.0668\).

Step 4: Finding Z-Scores for Diameters 0.0014 and 0.0026

Calculate Z-scores for both values. For \(X = 0.0014\), \[ Z = \frac{0.0014 - 0.002}{0.0004} = -1.5 \] and for \(X = 0.0026\), we already found \(Z = 1.5\) in Step 2.

Step 5: Probability Calculation for Diameter Between 0.0014 and 0.0026

Using the Z-scores, find the probability that Z is between -1.5 and 1.5. This is calculated as \(P(-1.5 < Z < 1.5) = P(Z < 1.5) - P(Z < -1.5)\). Looking up values: \(P(Z < 1.5) \approx 0.9332\) and \(P(Z < -1.5) \approx 0.0668\). So, \(P = 0.9332 - 0.0668 = 0.8664\).

Step 6: Adjusting Standard Deviation for Desired Probability

To find a standard deviation that yields a probability of 0.995 for Example b: use \(P(-z_1 < Z < z_2) = 0.995\). From Z-tables, find \(z_1 = -2.575\) and \(z_2 = 2.575\) for 0.995 probability. The new standard deviation \(\sigma'\) satisfies \[ X_1 = \mu + z_1 \sigma' = 0.0014 \text{ and } X_2 = \mu + z_2 \sigma' = 0.0026 \]. Solve the equations to find \(\sigma' = \frac{0.0026 - 0.0014}{2 \times 2.575}\).

Step 7: Compute New Standard Deviation

Calculate \(\sigma'\) from Step 6: \[ \sigma' = \frac{0.0026 - 0.0014}{2 \times 2.575} = \frac{0.0012}{5.15} \approx 0.000233 \].

Key Concepts

Z-Score

Understanding the Z-score can be really helpful when dealing with normal distributions. It's a way of expressing how far away a specific measurement is from the mean. By calculating a Z-score, we can determine how unusual or usual a particular value is, relative to the distribution's mean.

The Z-score can be found using the formula: \[ Z = \frac{X - \mu}{\sigma} \], where:

• \( X \) is the value of interest (e.g., dot diameter)
• \( \mu \) is the mean of the distribution
• \( \sigma \) is the standard deviation

For example, if the dot diameter is \(0.0026\), with a mean \(\mu = 0.002\) and standard deviation \(\sigma = 0.0004\), the Z-score is \(1.5\). This means that \(0.0026\) is 1.5 standard deviations above the mean.

Z-scores help us understand whether a value is typically expected or more rare based on the context of a normal distribution.

Probability Calculation

After finding a Z-score, you can easily calculate various probabilities. This involves using Z-score tables (also known as standard normal distribution tables) to find out how likely it is for a value to occur in a normal distribution.

For instance, if you want to know the probability of a dot diameter exceeding \(0.0026\) and calculate that the Z-score is \(1.5\), you will check the table for \(P(Z < 1.5)\). The Z-table gives us \(0.9332\), meaning there's a 93.32% chance a value is below 1.5 standard deviations from the mean. To find the probability of a diameter greater than \(0.0026\), we subtract this from 1 (as probabilities sum up to 1). Thus, \(1 - 0.9332 = 0.0668\), or 6.68% chance.

Similarly, to find the probability between two diameters, you compute Z-scores for both and then find respective probabilities from the Z-table, subtracting to get the desired probability. Such calculations open up understanding of data distribution and frequency of different outcomes.

Standard Deviation

The standard deviation is a key concept in statistics and normal distribution. It measures the spread or how much variation there is from the average (mean). In a normal distribution, smaller standard deviations mean that the values are closer to the mean, while larger ones indicate more spread.

When adjusting the standard deviation to achieve certain probabilities, as in exercise part (c), we're tweaking how spread out we expect our data to be to fit a new criterion. Let's say we need a probability of 0.995 for a specific range (like from 0.0014 to 0.0026). We adjust our standard deviation using the known value of Z, here approximately \(-2.575\) to \(2.575\). Solving the equation \[ \sigma' = \frac{X_2 - X_1}{2 \times (2.575)} \] where \(X_2\) and \(X_1\) are the limits of the interval gives us the new spread needed.

In our example, this calculation returns \(0.000233\), meaning the new standard deviation should be \(0.000233\) inches to achieve the desired probability coverage for the specified range. Understanding standard deviation allows us to comprehend not just individual events, but the behavior of entire data sets.