Question

Childhood lead poisoning is a public health concern in the United States. In a certain population, 1 child in 8 has a high blood lead level (defined as \(30 \mu \mathrm{g} / \mathrm{dl}\) or more). \(^{23}\) In a randomly chosen group of 16 children from the population, what is the probability that (a) none has high blood lead? (b) 1 has high blood lead? (c) 2 have high blood lead? (d) 3 or more have high blood lead? [Hint: Use parts (a)-(c) to answer part (d).]

Short answer

The probabilities are (a) none: \(\binom{16}{0} (1/8)^0 (7/8)^{16}\), (b) one: \(\binom{16}{1} (1/8)^1 (7/8)^{15}\), (c) two: \(\binom{16}{2} (1/8)^2 (7/8)^{14}\), (d) three or more: \(1 - (P(X=0)+P(X=1)+P(X=2))\).

Step-by-step solution

Understanding the problem

We are dealing with a problem of calculating probabilities of discrete outcomes where the probability of success (child with high blood lead level) is 1/8. Since we are selecting a group of 16 children, this scenario can be modeled as a binomial distribution with n = 16 trials and success probability p = 1/8.

Calculating probability for part (a)

The probability of none of the children having a high blood lead level can be calculated using the binomial probability formula, where the number of successes k = 0. The formula is: \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \] Substituting the values, we get: \[ P(X=0) = \binom{16}{0} \left(\frac{1}{8}\right)^0 \left(\frac{7}{8}\right)^{16} \]

Calculating probability for part (b)

To find the probability of exactly one child having a high blood lead level, we use the same binomial probability formula with k = 1. \[ P(X=1) = \binom{16}{1} \left(\frac{1}{8}\right)^1 \left(\frac{7}{8}\right)^{15} \]

Calculating probability for part (c)

To determine the probability of two children having a high blood lead level, we again apply the binomial formula with k = 2. \[ P(X=2) = \binom{16}{2} \left(\frac{1}{8}\right)^2 \left(\frac{7}{8}\right)^{14} \]

Determining probability for part (d)

The probability of 3 or more children having a high blood lead level is the complement of the probability of 0, 1, or 2 children having a high blood lead level. Thus, we calculate it as: \[ P(X \geq 3) = 1 - \left( P(X=0) + P(X=1) + P(X=2) \right) \]

Key concepts

Probability Calculations

Probability is an essential concept in mathematics that quantifies the likelihood of an event occurring. It ranges from 0, meaning the event will not occur, to 1, indicating that the event is certain to happen. In our exercise about childhood lead poisoning, probabilities are calculated using the formula for a binomial distribution.

The binomial distribution is a type of probability distribution with two possible outcomes, often termed as 'success' (event occurs) and 'failure' (event does not occur). Specifically, the formula used is: \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]

• \( n \) is the number of trials or attempts.
• \( k \) is the number of successes we are interested in.
• \( p \) is the probability of success on a single trial.

By plugging in the values corresponding to each scenario—none, one, or two children having high blood lead levels—into this formula, you can calculate the respective probabilities. This process is fundamental to making data-driven decisions in fields such as public health.

Discrete Random Variables

A discrete random variable is a variable that can take on a countable number of distinct outcomes. In the public health context of our example, the random variable \( X \) represents the number of children in a group of 16 with a high blood lead level—a number which can be 0, 1, 2, ..., up to 16.

When working with discrete random variables like this, each outcome has a specific probability associated with it. The probabilities of all possible outcomes of a discrete random variable must sum up to 1, reflecting certainty that one of the outcomes will occur.

Binomial Random Variable

Within the scope of discrete random variables, a binomial random variable is a special case relevant to this exercise. Here, the random variable \( X \), which counts the number of 'successes' (children with high blood lead levels), follows a binomial distribution because the number of trials (children tested) and the probability of 'success' (having a high blood lead level) are fixed.

Public Health Statistics

Public health statistics involve quantitative data related to the health of populations. These statistics often include incidence and prevalence of diseases, health behaviors, and the use of health care resources. In our exercise, the statistic that 1 in 8 children has a high blood lead level provides vital information for assessing and planning public health interventions.

By understanding the binomial distribution, public health officials can determine the probability of different numbers of children being affected by lead poisoning in a specific area. This facilitates risk assessment and resource allocation to prevent and treat lead exposure.

Importance of Accurate Estimates

Accurate probability calculations are crucial in this field to avoid underestimation or overestimation of health risks. For instance, estimating the likelihood of multiple cases of high blood lead levels in a population enables efficient planning for medical interventions, education, and policy initiatives aimed at mitigating such public health concerns.