Problem 111 - Applied Statistics and Probability for Engineers

Question

The number of (large) inclusions in cast iron follows a Poisson distribution with a mean of 2.5 per cubic millimeter. Approximate the following probabilities: (a) Determine the mean and standard deviation of the number of inclusions in a cubic centimeter \((\mathrm{cc})\). (b) Approximate the probability that fewer than 2600 inclusions occur in a cc. (c) Approximate the probability that more than 2400 inclusions occur in a cc. (d) Determine the mean number of inclusions per cubic millimeter such that the probability is approximately 0.9 that 500 or fewer inclusions occur in a cc.

Step-by-Step Solution

Answer

(a) Mean: 2500, SD: 50; (b) 0.9772; (c) 0.9772; (d) 445 inclusions/mm³.

Step 1: Determine Mean and Standard Deviation

First, we need to find the mean and standard deviation for a cubic centimeter (cc). Since there are 1000 cubic millimeters in a cubic centimeter, we multiply the mean per cubic millimeter by 1000. Thus, the mean number of inclusions per cc is \( ext{mean} = 2.5 \times 1000 = 2500 \). The standard deviation for a Poisson distribution is the square root of the mean, so \( ext{standard deviation} = \sqrt{2500} = 50 \).

Step 2: Approximate Probability of Fewer than 2600 Inclusions

To find the probability of fewer than 2600 inclusions, we use a normal approximation to the Poisson distribution. Therefore, we treat the distribution as approximately normal with mean 2500 and standard deviation 50. For \( P(X < 2600) \), we need to convert 2600 to a standard normal variable \( z \). Hence, \( z = \frac{2600 - 2500}{50} = 2 \). Using standard normal distribution tables, \( P(Z < 2) \approx 0.9772 \). So, the probability is approximately 0.9772.

Step 3: Approximate Probability of More than 2400 Inclusions

For \( P(X > 2400) \), first convert 2400 to a standard normal variable: \( z = \frac{2400 - 2500}{50} = -2 \). Using standard normal tables, \( P(Z < -2) \approx 0.0228 \). Hence, \( P(X > 2400) = 1 - P(X < 2400) = 1 - 0.0228 = 0.9772 \). Thus, the probability is approximately 0.9772.

Step 4: Mean for 500 or Fewer Inclusions with 0.9 Probability

We need to find a new mean \( \lambda \) such that \( P(X \leq 500) \approx 0.9 \). Using a normal approximation, this requires normalizing the variable \( X \) where \( z = \frac{500 - \lambda}{\sqrt{\lambda}} \approx 1.2816 \) (for a cumulative probability of 0.9). Solving \( \frac{500 - \lambda}{\sqrt{\lambda}} = 1.2816 \) will give us the required \( \lambda \). After solving, \( \lambda \approx 445 \) inclusions per cubic millimeter.

Key Concepts

Normal Approximation

When dealing with Poisson distributions with a large mean, it is often convenient to use the normal approximation for probability calculations. This is because the Poisson distribution tends to be skewed when the mean is small, but becomes more symmetric and approximates a normal distribution as the mean increases.

To do this, we consider the Poisson distribution as approximately normal, using the given mean and calculating the standard deviation accordingly. Normal approximation is a powerful tool because it lets us use the properties of the normal distribution to find probabilities that would otherwise be difficult to calculate directly from the Poisson distribution.

This method simplifies complex calculations and is especially useful when working with large datasets where the Poisson distribution is involved. In practice, the greater the mean, the closer the Poisson distribution will resemble a bell curve, making the normal approximation more accurate.

Standard Deviation

The standard deviation is a measure of how spread out the values in a distribution are around the mean. For a Poisson distribution, the standard deviation is simply the square root of the mean. This means that if you know the mean of a Poisson distribution, calculating the standard deviation is quite straightforward.

In our exercise, we determined that the mean number of inclusions in one cubic centimeter is 2500. Therefore, the standard deviation is \( \sqrt{2500} = 50 \).

Understanding the standard deviation helps us gauge the variability in the number of inclusions across different cubic centimeters. A higher standard deviation indicates that the inclusions are more spread out; whereas, a lower standard deviation means they are more clustered around the mean. Knowing the standard deviation is crucial for subsequent calculations, like translating the Poisson distribution into a normal one for probability approximation.

Probability Calculation

Calculating probabilities is a central task when working with statistical distributions. By normalizing our data, we use standard normal tables to find probabilities associated with events of interest.

In the exercise, we perform these calculations twice:

• For fewer than 2600 inclusions, we calculate a z-score and find that \( P(X < 2600) \approx 0.9772 \), which represents a high likelihood of having fewer than this many inclusions.
• For more than 2400 inclusions, we also calculate a z-score and use it to find that \( P(X > 2400) \approx 0.9772 \), indicating a high probability of exceeding this number of inclusions.

These calculations use the properties of both Poisson and normal distributions to make statistical predictions understandable and applicable for real-world scenarios.

Mean Value

The mean value in a Poisson distribution represents the average number of events occurring in a fixed interval. For example, in our problem, the mean number of large inclusions per cubic millimeter is 2.5.

To find the mean number of inclusions per cubic centimeter, which is a larger volume, multiply the original mean by the number of cubic millimeters in a cubic centimeter (1000). Thus, \( 2.5 \times 1000 = 2500 \) is the mean per cubic centimeter.

The mean is not only about central tendency but also plays a critical role in calculating the standard deviation and forming the foundation for probability approximations.

Additionally, we explore adjusting the mean to influence probability outcomes, such as ensuring a 0.9 probability for 500 or fewer inclusions. In this way, understanding and manipulating the mean is a key part of statistical strategy.