/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Choose a person aged 19 to 25 ye... [FREE SOLUTION] | 91影视

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Choose a person aged 19 to 25 years at random and ask, 鈥淚n the past seven days, how many times did you go to an exercise or fitness center or work out?鈥 Call the response Y for short. Based on a large sample survey, here is a probability model for the answer you will get: Days: 01234567 Probability: 0.68 0.05 0.07 0.08 0.05 0.04 0.01 0.02 (a) Show that this is a legitimate probability distribution. (b) Make a histogram of the probability distribution. Describe what you sec. (c) Describe the event \(Y<7\) in words. What is \(P(Y<7) ?\) (d) Express the event" "worked out at least once" in terms of Y. What is the probability of this event?

Short Answer

Expert verified
(a) Sum is 1. (b) Histogram shows majority at 0 days. (c) \(P(Y<7)=0.98\). (d) \(P(Y\geq1)=0.32\).

Step by step solution

01

Check Sum of Probabilities

To verify that the given probability distribution is legitimate, we need to ensure that the sum of all probabilities equals 1. Add the probabilities:\[0.68 + 0.05 + 0.07 + 0.08 + 0.05 + 0.04 + 0.01 + 0.02 = 1.00\]Since the sum is 1, this is a legitimate probability distribution.
02

Create Histogram

To create a histogram, plot the days (0 to 7) on the x-axis and the corresponding probabilities on the y-axis. Each bar represents a day with its height proportional to its probability: - **Day 0**: Height 0.68 - **Day 1**: Height 0.05 - **Day 2**: Height 0.07 - **Day 3**: Height 0.08 - **Day 4**: Height 0.05 - **Day 5**: Height 0.04 - **Day 6**: Height 0.01 - **Day 7**: Height 0.02 The histogram shows that working out 0 days is the most common response, while fewest people worked out for 6 or 7 days.
03

Describe Event Y

The event \(Y<7\) means that the person worked out less than 7 days in the past week, i.e., from 0 to 6 days. It's the sum of probabilities for these days:\[P(Y<7) = 0.68 + 0.05 + 0.07 + 0.08 + 0.05 + 0.04 + 0.01 = 0.98\]Thus, there is a 0.98 probability that a person worked out fewer than 7 days.
04

Describe "Worked Out At Least Once"

"Worked out at least once" refers to the event \(Y \geq 1\) (Y is 1 or greater). We calculate this probability by adding the probabilities for Y from 1 to 7:\[P(Y \geq 1) = 0.05 + 0.07 + 0.08 + 0.05 + 0.04 + 0.01 + 0.02 = 0.32\]This means there's a 0.32 probability that a person worked out at least once in the past week.

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Key Concepts

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

Histogram
Creating a histogram is a fantastic way to visualize data distribution and probability models. In a histogram, data is represented in the form of bars, where each bar's height reflects the frequency or probability of data points within a specific range. This makes it easier to interpret complex data sets.
In our exercise, the x-axis represents the number of days a person works out, ranging from 0 to 7. The y-axis represents the probability that corresponds with each day. Consequently, when plotting the histogram, we observe each bar representing different workout frequencies. For example, the tall bar for "Day 0" highlights that the majority didn't work out at all, with a probability of 0.68, showing it's the most common outcome, and smaller bars for "Days 6 and 7" indicate much less frequent exercise habits.
The visual layout of histograms helps spot the highest, lowest, and between variants in the data, offering an intuitive way of seeing how probabilities are spread across different outcomes.
Probability Model
A probability model is essential for predicting the likelihood of various outcomes within a random experiment. It comprises sets or lists of possible outcome probabilities, providing valuable insights into likely events.
In our scenario, the probability model indicates how often someone might visit an exercise center based on survey results. Each probability value specifies how likely each number of workout days (from 0 to 7) is. This allows us to calculate chances for specific events such as working out every day or not at all.
Understanding probability models aids in decision-making involving uncertain conditions. By having a structured model, we can better predict and interpret actual occurrences. It's a useful tool across many fields, such as finance, healthcare, and sports, helping adapt strategies to fit observed patterns.
Legitimate Probability Distribution
A legitimate probability distribution is a foundational concept in statistics and probability. For a probability distribution to be "legitimate," it must meet specific criteria. Firstly, the sum of all given probabilities must equal 1. This ensures that all possible outcomes are accounted for, totaling to a certainty that something will happen.
In the given exercise, we check this by summing the probabilities: \[0.68 + 0.05 + 0.07 + 0.08 + 0.05 + 0.04 + 0.01 + 0.02 = 1.00\]This confirms the probabilities form a legitimate distribution, mixing varied chances of working out from 0 to 7 times. Also, each individual probability must be between 0 and 1, reflecting realistic chances.
Verifying legitimacy is crucial as it ensures that data analysis and interpretations are based on accurate and complete probability models, leading to reliable predictions and insights.
Probability Calculation
Probability calculations are vital for understanding and interpreting statistical models. They enable us to derive meaningful probabilities for specific events. In this context, we calculate the probability for cases like working out less than 7 days or working out at least once.
For "working out less than seven days," which means from 0 to 6 days, combine probabilities accordingly:\[P(Y<7) = 0.68 + 0.05 + 0.07 + 0.08 + 0.05 + 0.04 + 0.01 = 0.98\]This means there's a 98% chance that someone won't work out every single day.
To figure out the probability of "working out at least once," we look at probabilities from 1 to 7 days:\[P(Y \geq 1) = 0.05 + 0.07 + 0.08 + 0.05 + 0.04 + 0.01 + 0.02 = 0.32\]So, there's a 32% chance of exercising at least one day.
These calculations form the basis for making informed guesses about behaviors and trends, offering quantitative assessments that facilitate strategic planning and understanding unobserved potential in data.

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Most popular questions from this chapter

Multiple choice: Select the best answer for Exercises 101 to 105. Joe reads that 1 out of 4 eggs contains salmonella bacteria. So he never uses more than 3 eggs in cooking. If eggs do or don't contain salmonella independently of each other, the number of contaminated eggs when Joe uses 3 chosen at random has the following distribution: (a) binomial; \(n=4\) and \(p=1 / 4\) (b) binomial; \(n=3\) and \(p=1 / 4\) (c) binomial; \(n=3\) and \(p=1 / 3\) (d) geometric; \(p=1 / 4\) (e) geometric; \(p=1 / 3\)

Geometric or not? Determine whether each of the following scenarios describes a geometric setting. If so, define an appropriate geometric random variable. (a) Shuffle a standard deck of playing cards well. Then turn over one card at a time from the top of the deck until you get an ace. (b) Lawrence is learning to shoot a bow and arrow. On any shot, he has about a 10\(\%\) chance of hitting the bull's-eye. Lawrence's instructor makes him keep shooting until he gets a bull's-eye.

Skee Ball Ana is a dedicated Skee Ball player (see photo) who always rolls for the 50 -point slot. The probability distribution of Ana's score \(X\) on a single roll of the ball is shown below. You can check that \(\mu_{X}=23.8\) and \(\sigma_{X}=12.63\) . Score: 10 20 30 40 50 Probability: 0.32 0.27 0.19 0.15 0.07 (a) A player receives one ticket from the game for every 10 points scored. Make a graph of the probability distribution for the random variable \(T=\) number of tickets Ana gets on a randomly selected throw. Describe its shape. (b) Find and interpret \(\mu_{T}\) . (c) Compute and interpret \(\sigma_{\mathrm{T}}\) .

Exercises 39 and 40 refer to the following setting. Ms. Hall gave her class a 10 -question multiple-choice quiz. Let \(X=\) the number of questions that a randomly selected student in the class answered correctly. The computer output below gives information about the probability distribution of \(X .\) To determine each student's grade on the quiz (out of \(100 ),\) Ms. Hall will multiply his or her number of correct answers by \(10 .\) Let \(G=\) the grade of a randomly chosen student in the class. N Mean Median StDev Min Max Q1 Q3 30 7.6 8.5 1.32 4 10 8 9 Easy quiz (a) Find the median of G. Show your method. (b) Find the \(I Q R\) of \(G\) . Show your method. (c) What shape would the probability distribution of G have? Justify your answer.

A sample survey contacted an SRS of 663 registered voters in Oregon shortly after an clection and asked respondents whether they had voted. Voter records show that 56\(\%\) of registered voters had actually voted. We will see later that in repeated random samples of size 663 , the proportion in the sample who voted (call this proportion With vary according to the Normal distribution with mean \(\mu=0.56\) and standard deviation \(\sigma=0.019\) (a) If the respondents answer truthfully, what is \(\mathrm{P}(0.52 \leq V \leq 0.60)\) ? This is the probability that the sample proportion \(V\) estimates the population proportion 0.56 within \(\pm 0.04\) (b) In fact, 72\(\%\) of the respondents said they had voted \((V=0.72) .\) If respondents answer truthfully, what is \(P(V \geq 0.72) ?\) This probability is so small that it is good evidence that some people who did not vote claimed that they did vote.

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