Problem 39 - A Graphical Approach to Precalculus with Limits

Question

The table gives the results of a survey of \(14,000\) college students who were cigarette smokers in a recent year. $$\begin{array}{|l|c|} \hline \begin{array}{l} \text { Number of Cigarettes } \\\ \text { per Day } \end{array} & \begin{array}{c} \text { Percent } \\\ \text { (as a decimal) } \end{array} \\ \hline \text { Less than } 1 & 0.45 \\\ 1 \text { to } 9 & 0.24 \\ 10 \text { to } 19 & 0.20 \\ \text { A pack of } 20 \text { or more } & 0.11 \end{array}$$ Using the percents as probabilinies, approximate the probability that, out of 10 of these shudent smokers selected at random, the following were true. Fewer than 2 smoked between 1 and 19 cigarettes per day.

Step-by-Step Solution

Answer

The probability that fewer than 2 out of 10 students smoke between 1 and 19 cigarettes per day is 0.062, or 6.2%.

Step 1: Understand the problem

We need to find the probability that fewer than 2 out of 10 students smoked between 1 and 19 cigarettes per day. This involves calculating the probability of an event occurring "fewer than" a certain number of times in a fixed number of trials, which suggests using a binomial probability distribution.

Step 2: Identify parameters for the binomial distribution

Define the parameters for the binomial distribution: Let \( n = 10 \) be the number of trials (students), and \( p \) be the probability of success (a student smoking between 1 and 19 cigarettes). This is given by the sum of the probabilities of smoking between 1 to 9 cigarettes (0.24) and 10 to 19 cigarettes (0.20), so \( p = 0.24 + 0.20 = 0.44 \).

Step 3: Compute required probabilities

We need to calculate the probability that fewer than 2 students (\( X < 2 \)) smoke between 1 and 19 cigarettes. This involves summing the probabilities for \( X = 0 \) and \( X = 1 \). The binomial probability formula is given by:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where \( n = 10 \) and \( k \) is the number of successes.

Step 4: Calculate for X = 0

Calculate \( P(X = 0) \):\[P(X = 0) = \binom{10}{0} (0.44)^0 (0.56)^{10} = 1 \times 1 \times 0.56^{10}\]Substitute \( 0.56^{10} \) and calculate the value.

Step 5: Calculate for X = 1

Calculate \( P(X = 1) \):\[P(X = 1) = \binom{10}{1} (0.44)^1 (0.56)^9 = 10 \times 0.44 \times 0.56^9\]Substitute the given values and calculate.

Step 6: Sum the probabilities

Sum the probabilities from Steps 4 and 5 to find \( P(X < 2) \):\[P(X < 2) = P(X = 0) + P(X = 1)\]Add the values calculated for \( P(X = 0) \) and \( P(X = 1) \).

Step 7: Conclusion

The calculated result is the probability that fewer than 2 out of 10 randomly chosen students smoked between 1 and 19 cigarettes per day.

Key Concepts

Probability Calculations

Probability calculations allow us to quantify the likelihood of different outcomes. They play a crucial role in decision-making and predicting future events based on past data. In the context of a binomial probability distribution, each trial has two possible outcomes: success or failure.

For example, in our problem, each student smoker either smokes between 1 and 19 cigarettes (a success) or doesn't (a failure). The probability of success in each individual trial is denoted by \( p \), while the complement, \( 1-p \), represents the probability of failure.

To calculate specific probabilities, we use the binomial probability formula:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where \( n \) is the number of trials, and \( k \) is the number of successful outcomes we're interested in. This formula gives us the probability of exactly \( k \) successes in \( n \) trials. This is fundamental to solving problems where there's a need to determine the likelihood of a specific number of successes, like fewer than 2 students meeting a certain criterion.

Discrete Probability Distributions

Discrete probability distributions deal with events that occur in countable sample spaces. Unlike continuous distributions, they handle outcomes that can be enumerated, making them ideal for counting occurrences in trials, such as flipping coins or checking cigarette consumption among students.

A discrete distribution assigns probabilities to each possible outcome, such that the sum of all probabilities equals 1. In the case of our problem, we're focusing on events where students fall into categories based on their smoking habits. Specifically, calculations involve determining the chance of a student choosing to smoke a certain number of cigarettes per day.

These probability assignments create a model of expected outcomes based on known data, helping us infer information about larger populations by examining samples. This kind of distribution is especially useful in surveys and experiments where outcomes must be explicitly categorized and recorded.

Binomial Theorem

The binomial theorem provides a powerful way to break down expressions that are raised to a power, applicable in various statistical methods, including binomial probability distributions. It is the theoretical foundation for calculations where we deal with two possible outcomes in repeated trials, like deciding whether students fall into a certain smoking category.

The distribution follows the pattern of expansion of \((a+b)^n\), where each term corresponds to different numbers of \(a\) and \(b\). For a binomial probability, \(a\) and \(b\) represent the probabilities of two possible outcomes, success and failure, and \( n \) is the number of trials.

The binomial theorem enables us to predict the probability distribution of outcomes over many trials by expanding the polynomial formula, allowing us to compute probabilities such as 'fewer than 2 successes.'

• The theorem supports numerical predictions based on binomial experiments.
• Efficiently handles calculations involving a fixed number of trials.
• Simplifies complex probability scenarios into manageable computations.

In the context of surveys or experiments, the binomial theorem efficiently aids in translating observed data into predictive insights.