Problem 38 - 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. Five smoked a pack or more per day.

Step-by-Step Solution

Answer

The probability that five of the students smoke a pack or more per day is approximately 0.000226.

Step 1: Identify Known Values

We are given that 10 students are selected. The probability that a single selected student smokes a pack or more per day is given by the table as 0.11.

Step 2: Define the Binomial Distribution

Since we are interested in the number of students (out of 10) who smoke a pack or more per day, this is a binomial distribution problem where the number of trials, \( n = 10 \), and the probability of success (smoking a pack or more) is \( p = 0.11 \).

Step 3: Use the Binomial Formula

The probability of exactly \( k \) successes (students who smoke a pack or more per day) in \( n \) trials is given by the binomial formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]For this problem, \( n = 10 \), \( k = 5 \), and \( p = 0.11 \).

Step 4: Calculate Combination

Calculate the combination \( \binom{10}{5} \), which represents the number of ways to choose 5 students from 10. \[ \binom{10}{5} = \frac{10!}{5!\,(10-5)!} = 252 \]

Step 5: Compute Probability Components

Calculate each part of the binomial formula separately:- \( p^k = (0.11)^5 \)- \((1-p)^{n-k} = (0.89)^5\)

Step 6: Plug Values into Formula

Plug the values into the binomial formula:\[ P(X = 5) = 252 \cdot (0.11)^5 \cdot (0.89)^5 \]

Step 7: Calculate Final Probability

Calculate the probability using the values computed: \[ (0.11)^5 \approx 1.6115 \times 10^{-5} \]\[ (0.89)^5 \approx 0.5584 \]\[ P(X=5) = 252 \cdot 1.6115 \times 10^{-5} \cdot 0.5584 \approx 0.000226 \]

Key Concepts

Probability Calculations

Probability calculations involve determining how likely an event is to occur. In the context of a binomial distribution problem, like determining how many students smoke a certain amount of cigarettes, probabilities define the chance of finding specific outcomes in repeated trials.

To calculate these probabilities, start by understanding the given chances. For example, if 11% (or 0.11 as a decimal) of students smoke a pack or more per day from the survey data, we use this as our basic probability for one student. This value sets the premise for further calculations.

The goal is to find out what are the chances of exactly a selected number doing the desired behavior in a group. Here, we use a method called a probability mass function specific for binomial distribution to calculate. This function accounts for the number of trials, the number of successes of interest, and the initial probability on one event.

The binomial distribution is useful because it provides a specific formula to calculate the probability of achieving a set number of successes across multiple independent trials, where the likelihood of success is constant for each trial. Understanding and using this formula can help predict outcomes in various practical scenarios.

Combination Formula

The combination formula is vital in the realm of probability and statistics when determining how many ways a particular event can happen. It finds key usage in solving problems dealing with categories like binomial distribution.

In our given problem, the combination formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!}\) is used to find out the number of ways to choose \( k \) successful events from \( n \) trials, disregarding the order of the outcomes. The '!' symbol denotes factorial and refers to multiplying a series of descending natural numbers.

Breaking it down further, \( n \) is the total number of events we examine, while \( k \) represents the number of successes we want from those events. For instance, when choosing 5 student smokers out of a total of 10, the formula \( \binom{10}{5} \) results in 252. This means there are 252 distinct ways to select 5 smokers from a group of 10 students.

This combination's role is to provide the number of ways a successful outcome can appear without considering order—essential for calculating probabilities like binomial distributions.

Statistics in Survey Analysis

In statistics, surveys are a valuable tool for collecting data about particular populations, like in the exercise involving 14,000 students. Analyzing survey data involves breaking down the collected responses to understand present patterns and trends.

When translating survey percentages into probabilities, treat these percentages as probabilities to understand likely future outcomes. For example, the survey data tells us 11% were students who smoked a pack or more daily, which we use directly in probabilistic models to predict broader population trends.

Using statistics in survey analysis similarly helps to simplify complex data into more manageable, understandable forms. This simplification allows for easier probabilistic modeling and inference. Through probability and binomial calculations, we can predict behaviors or tendencies based on the sample data, offering a practical overview of a designated population.

This conversion of raw survey data into meaningful statistical insights aids decision making and enhances understanding of specific population characteristics. It bridges the gap between data collection and actionable conclusions, effectively guiding future surveys and related research problems.