Problem 37
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. Four smoked fewer than 10 cigarettes per day.
Step-by-Step Solution
Verified Answer
The probability is approximately 0.042.
1Step 1: Identify Key Information
The table provides the percentages of smokers in various categories of cigarette consumption. For this problem, we focus on smokers who consume fewer than 10 cigarettes per day. This category includes those who smoke "Less than 1" and "1 to 9" cigarettes per day, which total to the probabilities of 0.45 and 0.24, respectively.
2Step 2: Calculate the Probability for Fewer Than 10 Cigarettes
The probability of a student smoking fewer than 10 cigarettes per day is the sum of the probabilities of the two relevant categories: \[ P( ext{fewer than 10}) = 0.45 + 0.24 = 0.69 \]
3Step 3: Set Up the Binomial Probability Scenario
We need to find the probability that exactly 4 out of 10 students smoke fewer than 10 cigarettes per day. This is a binomial probability problem where the number of trials is 10, the number of successes we want is 4, and the probability of success on each trial is 0.69.
4Step 4: Define the Binomial Probability Formula
The formula for the binomial probability of exactly \( k \) successes in \( n \) trials is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( n \) is the number of trials, \( k \) is the number of successes, and \( p \) is the probability of success.
5Step 5: Calculate the Binomial Probability
Substitute \( n = 10 \), \( k = 4 \), and \( p = 0.69 \) into the formula: \[ P(X = 4) = \binom{10}{4} (0.69)^4 (1-0.69)^6 \] Calculate \( \binom{10}{4} = 210 \), \( (0.69)^4 \approx 0.227 \), and \( (0.31)^6 \approx 0.00088 \). Thus, \[ P(X = 4) \approx 210 \times 0.227 \times 0.00088 \approx 0.042 \]
6Step 6: Interpret the Result
The probability that exactly 4 out of 10 selected student smokers smoke fewer than 10 cigarettes per day is approximately 0.042.
Key Concepts
ProbabilityBinomial DistributionStatistics Concepts
Probability
Probability is a way to quantify the likelihood of a particular outcome happening. It assigns a number between 0 and 1 to express this likelihood.
A probability of 0 means the event is impossible, while a probability of 1 means the event is certain. For events that actually could happen, like a student smoking fewer than 10 cigarettes, probabilities lie somewhere in between these extremes.
In this exercise, we calculate the probability of a student smoking fewer than 10 cigarettes per day, which is done by combining individual probabilities from two groups within the table. These were the groups smoking "Less than 1" and "1 to 9" cigarettes daily. Adding their probabilities gives us \(0.45 + 0.24 = 0.69\).
This probability represents the likelihood of a single trial, or instance, in our following exploration of binomial distribution.
A probability of 0 means the event is impossible, while a probability of 1 means the event is certain. For events that actually could happen, like a student smoking fewer than 10 cigarettes, probabilities lie somewhere in between these extremes.
In this exercise, we calculate the probability of a student smoking fewer than 10 cigarettes per day, which is done by combining individual probabilities from two groups within the table. These were the groups smoking "Less than 1" and "1 to 9" cigarettes daily. Adding their probabilities gives us \(0.45 + 0.24 = 0.69\).
This probability represents the likelihood of a single trial, or instance, in our following exploration of binomial distribution.
Binomial Distribution
The Binomial Distribution is a statistical method used to model the number of successes in a set number of independent trials. Each trial has the same probability of success.
Imagine rolling a die a certain number of times and wanting to know the probability of rolling a six a specific number of times.
In this exercise, we consider 10 randomly selected students who smoke, and how many among them smoke fewer than 10 cigarettes. We apply binomial distribution with the parameters: number of trials \(n = 10\), desired number of successes \(k = 4\), and the probability of success \(p = 0.69\).
The binomial distribution helps us to find the probability of obtaining exactly 4 successes, where a success is defined as a student smoking fewer than 10 cigarettes daily.
Imagine rolling a die a certain number of times and wanting to know the probability of rolling a six a specific number of times.
In this exercise, we consider 10 randomly selected students who smoke, and how many among them smoke fewer than 10 cigarettes. We apply binomial distribution with the parameters: number of trials \(n = 10\), desired number of successes \(k = 4\), and the probability of success \(p = 0.69\).
The binomial distribution helps us to find the probability of obtaining exactly 4 successes, where a success is defined as a student smoking fewer than 10 cigarettes daily.
Statistics Concepts
Understanding key statistics concepts is central to grasping this exercise. Binomial probability is just one piece of the larger puzzle we encounter in statistics.
When approaching this task, the first concept to understand is the use of probability as a measure. It's a foundation in determining outcomes in binomial scenarios.
Next is the concept of trials and successes in binomial distribution. Each trial is an event, like selecting one student, and each success meets a defined condition, such as smoking fewer than 10 cigarettes.
Lastly, calculating a probability like \(P(X = 4)\) requires comprehension of several steps: using the combination formula \(\binom{n}{k}\), raising \(p\) to the power of \(k\), and considering its complement \((1-p)\), all integral to predicting compound outcomes.
Such concepts connect significantly to real-life statistical analyses, equipping students with tools for data interpretation and informed decision-making.
When approaching this task, the first concept to understand is the use of probability as a measure. It's a foundation in determining outcomes in binomial scenarios.
Next is the concept of trials and successes in binomial distribution. Each trial is an event, like selecting one student, and each success meets a defined condition, such as smoking fewer than 10 cigarettes.
Lastly, calculating a probability like \(P(X = 4)\) requires comprehension of several steps: using the combination formula \(\binom{n}{k}\), raising \(p\) to the power of \(k\), and considering its complement \((1-p)\), all integral to predicting compound outcomes.
Such concepts connect significantly to real-life statistical analyses, equipping students with tools for data interpretation and informed decision-making.
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