Problem 197 - Applied Statistics and Probability for Engineers

Question

The percentage of people exposed to a bacteria who become ill is \(20 \%\). Assume that people are independent. Assume that 1000 people are exposed to the bacteria. Approximate each of the following: (a) Probability that more than 225 become ill (b) Probability that between 175 and 225 become ill (c) Value such that the probability that the number of people who become ill exceeds the value is 0.01

Step-by-Step Solution

Answer

(a) 0.0222 (b) 0.9361 (c) 229.5

Step 1: Understand the Problem

We are given that 20% of people exposed to a bacteria become ill. We are to assume that 1000 people are exposed and need to find probabilities related to how many of them become ill.

Step 2: Identify the Distribution

This problem involves a binomial distribution because each person exposed either becomes ill or doesn't, independently of others. The parameters for this distribution are: number of trials, \( n = 1000 \), and probability of success, \( p = 0.2 \).

Step 3: Approximate with Normal Distribution

We use the normal approximation to the binomial distribution. The mean \( \mu \) is given by \( n \times p = 1000 \times 0.2 = 200 \). The variance \( \sigma^2 \) is \( n \times p \times (1-p) = 1000 \times 0.2 \times 0.8 = 160 \), giving a standard deviation \( \sigma = \sqrt{160} \approx 12.65 \).

Step 4: Compute (a): Probability that more than 225 become ill

Convert the binomial problem to a normal one. Use the continuity correction to find \( P(X > 225.5) \). Calculate the z-score: \( z = \frac{225.5 - 200}{12.65} \approx 2.01 \). Use the z-table to find \( P(Z > 2.01) \), which is approximately 0.0222.

Step 5: Compute (b): Probability that between 175 and 225 become ill

Apply the continuity correction: \( P(174.5 < X < 225.5) \). Compute the z-scores: \( z_1 = \frac{174.5 - 200}{12.65} \approx -2.02 \) and \( z_2 = \frac{225.5 - 200}{12.65} \approx 2.01 \). Use z-table values: \( P(-2.02 < Z < 2.01) = P(Z < 2.01) - P(Z < -2.02) \), which is approximately 0.9578 - 0.0217 = 0.9361.

Step 6: Compute (c): Find value such that probability exceeds it is 0.01

We need \( P(X > k) = 0.01 \), so \( P(X \leq k) = 0.99 \). Find the z-score for 0.99, which is 2.33. Use \( z = \frac{k - 200}{12.65} = 2.33 \) to solve for \( k \): \( k = 2.33 \times 12.65 + 200 \approx 229.5 \).

Key Concepts

Normal Approximation

The binomial distribution is often used to model scenarios where you have a fixed number of trials and two outcomes, such as becoming ill or not from bacteria exposure. However, calculating probabilities directly from the binomial distribution can be complex for large numbers like 1000 trials. This is where the normal approximation becomes particularly useful.

When the sample size is sufficiently large, the binomial distribution can be approximated by a normal distribution. This is applicable to our problem because we are dealing with a scenario where 1000 individuals are exposed to a bacteria with a 20% infection rate. To use the normal approximation, we calculate the mean \( \mu \) and the standard deviation \( \sigma \) of the binomial distribution.

From our problem, the mean \( \mu \) is calculated as \( n \times p = 1000 \times 0.2 = 200 \), and the standard deviation \( \sigma \) is \( \sqrt{n \times p \times (1-p)} = \sqrt{1000 \times 0.2 \times 0.8} \approx 12.65 \). With these parameters, we can transform our binomial distribution into a normal one, making it easier to compute probabilities.

Continuity Correction

Continuity correction is an essential step when using the normal approximation for a binomial distribution. This is because the binomial distribution is discrete, while the normal distribution is continuous.

When converting a discrete variable to a continuous one, we adjust our calculations slightly, known as the continuity correction. For example, if we want to calculate the probability of more than 225 people becoming ill, we actually calculate \( P(X > 225.5) \). This small adjustment of adding or subtracting 0.5 is necessary for accuracy.

In problems where you're dealing with exact values like 'between 175 and 225', you would apply corrections such as \( P(174.5 < X < 225.5) \). These corrections help align the intervals to better fit a continuous distribution, improving the accuracy of your probability calculations.

Z-score

The z-score is a pivotal concept when utilizing the normal approximation. It allows you to transform normal distribution probabilities into standard normal distribution probabilities, which can easily be found in z-tables.

To calculate a z-score, you use the formula: \[ z = \frac{X - \mu}{\sigma} \] where \( X \) is the value you're interested in, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

In the bacteria problem, to determine the likelihood of more than 225 people becoming ill, we compute the z-score for 225.5 using this formula, resulting in \( z \approx 2.01 \).

Once you have the z-score, you can look up standard normal distribution tables to find the corresponding probability. A similar process is used for determining probabilities between values or finding a specific value with given probabilities, as shown in the exercise's solutions.

Probability Calculation

After understanding how to apply normal approximation, continuity correction, and compute z-scores, you can calculate probabilities. You can determine the likelihood of observing certain outcomes in a binomially-distributed scenario using these steps.

For instance, to find the probability of more than 225 people being ill, we calculate \( P(Z > 2.01) \) using the z-score of 2.01. From z-tables, \( P(Z > 2.01) \approx 0.0222 \), indicating there's about a 2.22% chance this occurs.

Similarly, for determining the probability of outcomes falling between two values, like between 175 and 225 ill people, we find the relevant z-scores, then use these to calculate the cumulative probabilities. Subtracting these probabilities gives the desired result.

Probability calculations using these methods increase precision, especially for large sample sizes, making them indispensable in simplifying complex binomial problems.