Problem 40
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. No more than 3 smoked less than 1 cigarette per day.
Step-by-Step Solution
Verified Answer
The probability is approximately 0.075.
1Step 1: Identify the Probability
The probability of a student smoking less than 1 cigarette per day is given as 0.45. Therefore, we set \( p = 0.45 \) for our calculations.
2Step 2: Define the Binomial Distribution
The situation can be modeled using a binomial distribution since we are dealing with a fixed number of trials (10 students), two possible outcomes (smoke less than 1 cigarette or not), and a constant probability of success (0.45). We denote this distribution as \( X \sim \text{Binomial}(10, 0.45) \).
3Step 3: Calculate the Desired Probability
We need to calculate the probability that no more than 3 students smoke less than 1 cigarette per day. Mathematically, this is written as \( P(X \leq 3) \). We sum the probabilities of getting 0, 1, 2, and 3 successes in a binomial distribution with parameters \( n = 10 \) and \( p = 0.45 \). This is calculated as: \[ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \]
4Step 4: Use the Binomial Probability Formula
The binomial probability formula is \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \( \binom{n}{k} \) is the number of combinations of \( n \) items taken \( k \) at a time. Apply this formula for \( k = 0, 1, 2, \) and \( 3 \).
5Step 5: Compute Each Probability
Calculate each term: - \( P(X = 0) = \binom{10}{0} (0.45)^0 (0.55)^{10} \)- \( P(X = 1) = \binom{10}{1} (0.45)^1 (0.55)^9 \)- \( P(X = 2) = \binom{10}{2} (0.45)^2 (0.55)^8 \)- \( P(X = 3) = \binom{10}{3} (0.45)^3 (0.55)^7 \)
6Step 6: Summing Up the Probabilities
Add the results of the probabilities calculated in Step 5 to find \( P(X \leq 3) \). This gives us the probability that no more than 3 students smoke less than 1 cigarette.
7Step 7: Final Solution Calculation
Use a calculator or statistical software to evaluate: - \( P(X = 0) \approx 0.0003 \)- \( P(X = 1) \approx 0.0033 \)- \( P(X = 2) \approx 0.0175 \)- \( P(X = 3) \approx 0.0537 \)Summing these: \( P(X \leq 3) \approx 0.0748 \)
Key Concepts
Probability CalculationProbability TheoryCombinatorics
Probability Calculation
Probability calculation involves determining the likelihood of specific outcomes in a given scenario. In the context of the binomial distribution, we calculate probabilities by considering specific trials and outcomes. Here, each student smoker is considered an independent trial with a fixed probability of success, which means they smoke less than 1 cigarette per day.
When we say "probability that no more than 3 students smoke less than 1 cigarette," we're summing up the probabilities of 0 to 3 students being smokers at the defined rate. 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 (students), \( p \) is the probability of success on any given trial, and \( k \) is the number of successful outcomes.
The probabilities are then combined, offering a complete picture of less than 1 cigarette/day smoking among a group of students. Understanding how to calculate and interpret these probabilities is crucial in data analysis and decision-making.
When we say "probability that no more than 3 students smoke less than 1 cigarette," we're summing up the probabilities of 0 to 3 students being smokers at the defined rate. 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 (students), \( p \) is the probability of success on any given trial, and \( k \) is the number of successful outcomes.
The probabilities are then combined, offering a complete picture of less than 1 cigarette/day smoking among a group of students. Understanding how to calculate and interpret these probabilities is crucial in data analysis and decision-making.
Probability Theory
Probability Theory is the branch of mathematics that deals with the analysis of random events. It is the foundation of statistical analysis and decision science. In our exercise, it facilitates understanding random behavior by modeling it through a probability distribution, specifically the binomial distribution.
In probability theory:
In probability theory:
- An event's probability measures its chance of happening.
- A probability closer to 1 means an event is more likely to happen, whereas a probability closer to 0 means it's less likely.
- The sum of probabilities of all possible outcomes of a given experiment equals 1.
Combinatorics
Combinatorics is a field of mathematics focused on counting, arranging, and finding patterns in sets. It plays a critical role in the binomial probability formula by allowing us to determine the number of ways events can occur. This is expressed through combinations, which appear as \( \binom{n}{k} \), the binomial coefficient, in our calculations.
In the formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \),
In the formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \),
- \( n! \) (n factorial) is the product of all positive integers up to \( n \).
- \( k! \) is the product of integers up to \( k \).
- \( (n-k)! \) is the product of integers up to \( n-k \).
Other exercises in this chapter
Problem 39
Write the indicated tem of each binomial expansion. Sixth term of \((4 h-j)^{8}\)
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Find the sum of the first 10 terms of each arithmetic sequence. $$a_{2}=9, a_{4}=13$$
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Use the fundamental principle of counting or permutations to solve each problem. A concert to raise money for an economics prize is to consist of 5 works: 2 ove
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