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91Ó°ÊÓ

College smokers. At a university, \(13 \%\) of students smoke. (a) Calculate the expected number of smokers in a random sample of 100 students from this university. (b) The university gym opens at 9 am on Saturday mornings. One Saturday morning at 8: 55 am there are 27 students outside the gym waiting for it to open. Should you use the same approach from part (a) to calculate the expected number of smokers among these 27 students?

Short Answer

Expert verified
(a) 13 smokers. (b) No, the approach might not apply.

Step by step solution

01

Identify Given Information

We are given that the probability of a student smoking at the university is \( p = 0.13 \). We need to calculate the expected number of smokers in a sample of 100 students.
02

Set Up the Equation for Expected Value

The expected value for a number of smokers in a sample can be calculated using the formula: \( E(X) = n \cdot p \). Here, \( n = 100 \) and \( p = 0.13 \).
03

Perform the Calculation

Substitute the values into the formula: \( E(X) = 100 \times 0.13 = 13 \). Therefore, the expected number of smokers in a sample of 100 students is 13.
04

Analyze the Second Scenario

For part (b), the situation involves a subset of students (27 at the gym) not randomly chosen. Without additional information about this subset, we cannot assume they represent the entire student body, so a different approach is necessary.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
When discussing probability, the concept of "Expected Value" is crucial. It's a statistical measure that gives us an average outcome for a random event over many repetitions. In simpler terms, if you conducted an experiment several times, the expected value would be the average result you would anticipate.
For example, consider the exercise from the university scenario. We know that 13% of students at the university are smokers. If we take a random sample of 100 students, we can calculate the expected number of smokers by multiplying the probability of smoking (\( p = 0.13 \)) by the number of students (\( n = 100 \)).
  • The calculation goes as follows: \( E(X) = n \times p = 100 \times 0.13 = 13 \).
So, on average, we expect to find 13 smokers in a random sample of 100 students. Keep in mind this is an average over many samples of 100 students.
Random Sample
A "Random Sample" refers to a group of subjects randomly chosen from a larger population. The randomness ensures each member of the population has an equal chance of being selected, providing a truthful representation of the entire group.
This is important because it allows statistical methods to work properly, ensuring the conclusions drawn from the sample are valid for the whole population.
  • For instance, in our university example, the sample of 100 students was randomly chosen from all university students, creating a fair representation of the student body regarding the smoking habit.
Using a random sample avoids biases that might skew the results. Thus, conclusions about the entire population can be drawn more accurately from a random sample.
University Students
Consider the scenario involving university students. This context provides a real-world setting for understanding statistical concepts like expected value and random sampling.
Students are often studied in statistical exercises because their population dynamics can illustrate many statistical principles.
  • For example, by knowing that 13% of university students smoke, we can practice calculating expected values for random samples of different sizes.
Moreover, understanding specific groups, like university students, highlights the necessity of random sampling to draw accurate conclusions about behavior across an entire population. Studying university students is not only practical but also helps connect theoretical statistical concepts to tangible real-world phenomena.
Binomial Distribution
The "Binomial Distribution" is a probability distribution that summarizes the probability of achieving a fixed number of successes in a specified number of trials.It is essential in calculating probabilities for events with two possible outcomes, like "success" or "failure," as seen in smoking or not smoking.
  • In the case of our exercise, each student represents a trial, with smoking being the success condition.
  • The number of smokers in a sample of 100 students follows a binomial distribution with parameters \( n = 100 \) and \( p = 0.13 \).
This distribution helps us calculate probabilities beyond just the expected value. For example, it can inform us about the likelihood of finding exactly 10 smokers in a sample of 100 or more than 15 smokers. Using the binomial distribution further enhances our understanding of variability and expectation in random sampling scenarios.

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Most popular questions from this chapter

Ice cream usually comes in 1.5 quart boxes (48 fluid ounces), and ice cream scoops hold about 2 ounces. However, there is some variability in the amount of ice cream in a box as well as the amount of ice cream scooped out. We represent the amount of ice cream in the box as \(X\) and the amount scooped out as \(Y\). Suppose these random variables have the following means, standard deviations, and variances: $$\begin{array}{cccc} \hline & \text { mean } & \text { SD } & \text { variance } \\ \hline X & 48 & 1 & 1 \\ Y & 2 & 0.25 & 0.0625 \\ \hline\end{array}$$ (a) An entire box of ice cream, plus 3 scoops from a second box is served at a party. How much ice cream do you expect to have been served at this party? What is the standard deviation of the amount of ice cream served? (b) How much ice cream would you expect to be left in the box after scooping out one scoop of ice cream? That is, find the expected value of \(X-Y\). What is the standard deviation of the amount left in the box? (c) Using the context of this exercise, explain why we add variances when we subtract one random variable from another.

Suppose \(80 \%\) of people like peanut butter, \(89 \%\) like jelly, and \(78 \%\) like both. Given that a randomly sampled person likes peanut butter, what's the probability that he also likes jelly?

The Behavioral Risk Factor Surveillance System (BRFSS) is an annual telephone survey designed to identify risk factors in the adult population and report emerging health trends. The following table displays the distribution of health status of respondents to this survey (excellent, very good, good, fair, poor) and whether or not they have health insurance. $$\begin{array}{rrrrrrr} & {\text { Health Status }} & \\ & \text { Excellent } & \text { Very good } & \text { Good } & \text { Fair } & \text { Poor } & \text { Total } \\ \hline \text { No } & 0.0230 & 0.0364 & 0.0427 & 0.0192 & 0.0050 & 0.1262 \\ \text { Yes } & 0.2099 & 0.3123 & 0.2410 & 0.0817 & 0.0289 & 0.8738 \\ \hline \text { Total } & 0.2329 & 0.3486 & 0.2838 & 0.1009 & 0.0338 & 1.0000 \end{array}$$ (a) Are being in excellent health and having health coverage mutually exclusive? (b) What is the probability that a randomly chosen individual has excellent health? (c) What is the probability that a randomly chosen individual has excellent health given that he has health coverage? (d) What is the probability that a randomly chosen individual has excellent health given that he doesn't have health coverage? (e) Do having excellent health and having health coverage appear to be independent?

Imagine you have a bag containing 5 red, 3 blue, and 2 orange chips. (a) Suppose you draw a chip and it is blue. If drawing without replacement, what is the probability the next is also blue? (b) Suppose you draw a chip and it is orange, and then you draw a second chip without replacement. What is the probability this second chip is blue? (c) If drawing without replacement, what is the probability of drawing two blue chips in a row? (d) When drawing without replacement, are the draws independent? Explain.

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