Question

The probability that a person will develop the flu after getting a flu shot is 0.01 . In a random sample of 200 people in a community who got a flu shot, what is the probability that 5 or more of the 200 people will get the flu? Use the Poisson approximation to binomial probabilities to find your answer.

Short answer

Answer: The probability that 5 or more people in the random sample of 200 will develop the flu after getting the flu shot is approximately 0.0527 or 5.27%.

Step-by-step solution

Identify given values and compute lambda (λ)

We are given the probability of an individual developing the flu after getting a flu shot (p = 0.01) and the sample size (n = 200). We first need to calculate the average number of occurrences (λ). This can be found by multiplying the probability (p) and the sample size (n): λ = n * p

Calculate λ

Now, calculate λ using the given values: λ = 200 * 0.01 λ = 2

Probability using Poisson approximation

The probability of having exactly k occurrences using Poisson approximation can be found using the formula: P(k) = e^(-λ) * (λ^k) / k! Where e is the base of the natural logarithm (approximately 2.71828), k is the number of occurrences, and λ is the average number of occurrences (calculated in Step 2). To find the probability of having 5 or more occurrences, we will first find the probability of having 0 to 4 occurrences (k = 0 to 4) using the above formula and then subtract the sum of these probabilities from 1.

Calculate probabilities for k = 0 to 4

Apply the Poisson formula for k = 0 to 4: P(0) = e^(-2) * (2^0) / 0! ≈ 0.1353 P(1) = e^(-2) * (2^1) / 1! ≈ 0.2707 P(2) = e^(-2) * (2^2) / 2! ≈ 0.2707 P(3) = e^(-2) * (2^3) / 3! ≈ 0.1804 P(4) = e^(-2) * (2^4) / 4! ≈ 0.0902

Calculate the probability for 5 or more occurrences

Now, calculate the sum of probabilities for k = 0 to 4: P_sum = P(0) + P(1) + P(2) + P(3) + P(4) ≈ 0.1353 + 0.2707 + 0.2707 + 0.1804 + 0.0902 ≈ 0.9473 To find the probability of 5 or more occurrences, subtract the sum of probabilities for k = 0 to 4 (P_sum) from 1: P(5 or more) = 1 - P_sum ≈ 1 - 0.9473 ≈ 0.0527 So, the probability that 5 or more people in the random sample of 200 will develop the flu after getting the flu shot is approximately 0.0527 or 5.27%.

Key concepts

Probability and Statistics

Probability and statistics are branches of mathematics that deal with the laws governing random events. The foundation of probability is the likelihood of a certain event occurring among a set of possible outcomes. In our day-to-day lives, we might come across the use of probability when we talk about chances or risks, like the chance of rain on a certain day or the risk of an investment.

Statistics, on the other hand, involves collecting, analyzing, interpreting, presenting, and organizing data. It helps us make sense of numerical data, identify patterns, and make inferences about populations based on samples. In the context of our exercise, probability helps us understand the chance of individuals developing the flu after vaccination, while statistics allows us to use a sample (200 people) to make an inference about the larger population.

Poisson Distribution

The Poisson distribution is a probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events happen with a known constant mean rate and independently of the time since the last event. This distribution is suitable for modeling the number of events in fixed intervals such as the number of emails received in an hour, the number of cars passing through a checkpoint, or, in our case, the number of people contracting the flu after vaccination in a sample of 200.

Mathematically, the Poisson distribution is defined by the parameter \( \lambda \), which is the average number of events in the interval. Key characteristics of the Poisson distribution include that it's discrete, and the average rate (or mean) of events is consistent throughout the observation period. The formula to calculate the probability of observing exactly \( k \) events is: \[ P(k) = \frac{e^{-\lambda} \lambda^k}{k!} \. The exercise demonstrates this using \( \lambda = 2 \) as the expected number of flu cases after vaccinating a sample of 200 people with a probability of 0.01, and computing probabilities for various numbers of actual occurrences.

Binomial Probability

The binomial probability distribution represents the probability of obtaining exactly \( k \) successes out of \( n \) trials in a sequence of independent experiments, where each trial is known as a Bernoulli trial and has only two possible outcomes: success or failure. This distribution is characterized by two parameters: the number of trials \( n \) and the probability of success on a single trial \( p \).

The probability mass function (PMF) for binomial probabilities is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \: where \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) successes from \( n \) trials.

In practice, when \( n \) is large or \( p \) is very small, calculations using the binomial formula can be cumbersome. In such cases, the Poisson distribution serves as a good approximation to the binomial distribution if \( np \) (mean number of successes) is small compared to \( n \) and can thus simplify the computational process, as demonstrated in our exercise.