Problem 109 - Applied Statistics and Probability for Engineers

Question

Among homeowners in a metropolitan area, \(75 \%\) recycle plastic bottles each week. A waste management company services 1500 homeowners (assumed independent). Approximate the following probabilities: (a) At least 1150 recycle plastic bottles in a week (b) Between 1075 and 1175 recycle plastic bottles in a week

Step-by-Step Solution

Answer

(a) \(0.1251\), (b) \(0.9784\)

Step 1: Identify the Distribution

The problem involves a large sample of homeowners with a probability of success across each trial. This indicates a binomial distribution with parameters \(n = 1500\) and \(p = 0.75\).

Step 2: Use Normal Approximation

Given the large \(n\), the binomial distribution can be approximated using a normal distribution. Calculate the mean \( \mu \) and the standard deviation \( \sigma \) using the formulas: \( \mu = np = 1500 \times 0.75 = 1125 \) and \( \sigma = \sqrt{np(1-p)} = \sqrt{1500 \times 0.75 \times 0.25} \approx 21.65 \).

Step 3: Calculating Probability for Part (a)

Convert the binomial probability to a normal probability for \(X \geq 1150\). Compute the z-score: \( z = \frac{1150 - 1125}{21.65} \approx 1.15 \). Using the standard normal table or calculator, find \( P(Z \geq 1.15) \), which is approximately \(0.1251\).

Step 4: Calculating Probability for Part (b)

Determine the probability for \(1075 \leq X \leq 1175\). Calculate the z-scores: \( z_{1} = \frac{1075 - 1125}{21.65} \approx -2.31 \) and \( z_{2} = \frac{1175 - 1125}{21.65} \approx 2.31 \). Use the standard normal table to find the probabilities: \( P(-2.31 \leq Z \leq 2.31) \), which results in approximately \(0.9784\).

Key Concepts

Binomial Distribution

The binomial distribution is an essential concept in statistics. It represents the number of successes in a fixed number of independent, identical experiments, each with the same probability of success.

• In the context of our exercise, a "success" is defined as a homeowner recycling plastic bottles in a given week.
• The total number of trials, or homeowners, is 1500.
• The probability of success, or the fraction of homeowners recycling each week, is 0.75.

Understanding the binomial distribution helps in estimating real-world scenarios where similar trials are repeated multiple times. It allows us to calculate probabilities related to the number of successes over these trials. This distribution applies only when each trial is independent, and the probability of success remains constant.
In our exercise, the large sample size allows us to approximate the binomial distribution with a normal distribution to simplify computations.

Probability

Probability quantifies the likelihood of an event occurring. For any possible event, probability values range between 0 (impossible event) and 1 (certain event). In terms of binomial distribution, we focus on finding the probability that a certain number of events (e.g., homeowners recycling) occurs.

• In Case (a), we are interested in finding the probability of at least 1150 homeowners recycling among 1500.
• In Case (b), we want the probability that between 1075 and 1175 homeowners recycle.

With a large sample size, using tables or calculators to determine the probability becomes easier when the binomial distribution is approximated to a normal distribution.
This approximation lets us use z-scores, which are standardized scores representing how far away a data point is from the mean in terms of standard deviations. The calculation of these probabilities requires converting our counts into continuous values so we can utilize the standard normal distribution, which simplifies complex computations.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values.

• It helps us understand how much the values in our data set deviate from the mean.
• A small standard deviation signifies that the data points are close to the mean.
• A large standard deviation indicates that the data points are spread out over a wider range.

For a binomial distribution, the standard deviation is calculated with the formula:
\[ \sigma = \sqrt{np(1-p)} \]
In the exercise, we calculated the standard deviation to be approximately 21.65.
This number tells us how much variability we can expect in the number of homeowners recycling plastic bottles each week around the mean. By understanding standard deviation, we can better interpret the data's consistency and predict the likelihood of certain outcomes over a large number of trials. It is crucial for forming the basis of normal approximation, allowing us to assess the probability of outcomes using the z-score.