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Americans are really getting away while on vacation. In fact, among small business owners, more than half \((51 \%)\) say they check in with the office at least once a day while on vacation; only \(27 \%\) say they cut the cord completely. \({ }^{7}\) If 20 small business owners are randomly selected, and we assume that exactly half check in with the office at least once a day, then \(n=20\) and \(p=.5 .\) Find the following probabilities. a. Exactly 16 say that they check in with the office at least once a day while on vacation. b. Between 15 and 18 (inclusive) say they check in with the office at least once a day while on vacation. c. Five or fewer say that they check in with the office at least once a day while on vacation. Would this be an unlikely occurrence?

Short Answer

Expert verified
Answer: The probability that five or fewer small business owners check in with the office at least once a day while on vacation is approximately 0.0207. Since the value is close to 0, it means the probability of this happening is low, so we can say it's an unlikely occurrence.

Step by step solution

01

Understand the binomial distribution formula

The binomial distribution formula is given as \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\), where \(n\) is the sample size, \(k\) is the number of successes, and \(p\) is the probability of success.
02

Use the binomial distribution formula to find the probability for part a

For part (a), we need to find the probability that exactly 16 small business owners check in with the office at least once a day while on vacation. Here, \(n=20\) and \(k=16\). Plugging these values into the binomial formula, we have: \(P(X=16) = \binom{20}{16} (0.5)^{16} (1-0.5)^{20-16} = \binom{20}{16} (0.5)^{16} (0.5)^{4}\) Now, calculate and find the probability:\(P(X=16) \approx 0.0577\)
03

Use the binomial distribution formula to find the probability for part b

For part (b), we need to find the probability that between 15 and 18 (inclusive) small business owners check in with the office at least once a day while on vacation. Here, we need to find the sum of the probabilities for \(k = 15, 16,17,\) and \(18\): \(P(15 \leq X \leq 18) = P(X=15) + P(X=16) + P(X=17) + P(X=18)\) Now, use the binomial formula to find probabilities for each value of \(k\) and sum them up: \(P(15 \leq X \leq 18) \approx 0.2026\)
04

Use the binomial distribution formula to find the probability for part c

For part (c), we need to find the probability that five or fewer small business owners check in with the office at least once a day while on vacation. Here, we need to find the sum of probabilities for \(k = 0,1,2,3,4,\) and \(5\): \(P(X \leq 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)\) Now, use the binomial formula to find probabilities for each value of \(k\) and sum them up: \(P(X \leq 5) \approx 0.0207\) Now, we need to determine if this occurrence is unlikely. Since the value is close to 0, it means the probability of this happening is low, so we can say it's an unlikely occurrence.

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Key Concepts

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

Probability Calculation
Probability plays a crucial role in understanding and predicting various phenomena. In this scenario, it helps us determine how often certain vacation work habits occur among small business owners. The concept of calculating probabilities involves determining how likely an actual event is to happen. When we talk about binomial distribution in probability calculation, we're dealing with a specific type of distribution based on binary outcomes, such as success or failure.

In binomial probability, we use a formula to calculate the likelihood of a given number of successes happening in a series of trials. Here's the formula:
  • \( P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \)
Where:
  • \( n \) is the number of trials or sample size, which in this example is 20.
  • \( k \) is the number of times the event of interest occurs, which varies in the sub-questions.
  • \( p \) is the probability of success on any given trial, which here is set to 0.5.
When small business owners are surveyed, calculations are made regarding the number who maintain work connections during vacations using this formula. Each sub-part of the problem calculates the probability for different values of \( k \), helping us assess specific vacation work habits.
Small Business Owners
Small business owners play a fundamental role in the economy. With their ventures, they bring innovation, create jobs, and often work beyond regular hours. Within this context, their work habits, even when on vacation, become a topic of interest. Understanding how small business owners balance work and leisure can offer insights into stress levels, productivity, and the broader dynamics of managing a business.
  • More than half check in with the office daily during vacations, showcasing their strong attachment and responsibility to their businesses.
  • Only 27% cut ties completely when on vacation, reflecting on the nature of their commitment.
When we use statistical tools like binomial distribution to study small business owners, we delve into behaviors and expectations surrounding work-life balance. These results could affect how resources or support are provided to them, acknowledging the pressures they face while also valuing their downtime.
Vacation Work Habits
Vacation work habits refer to how individuals, especially professionals like small business owners, manage their work responsibilities while taking time off. Despite the purpose of vacations being to relax and disconnect, modern communication tools often blur those lines.
This exercise illustrates the likelihood of different vacation work habits occurring among small business owners. Some key observations:
  • Many choose to remain partially connected, with 51% checking in at least once a day.
  • This trend highlights a potential inability to fully disconnect, possibly due to technology or work culture pressures.
  • A smaller fraction, 27%, manage to abstain completely from checking in.
Using probability calculations like in the exercise, we identify how common non-disconnection is among vacationers. Analyzing cases where up to five out of the 20 business owners stick to their work routines suggests it's rare, hence labeled unlikely. Such conclusions assist in understanding owner behaviors and might prompt reassessments of how vacations are perceived and taken by small business owners.

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Most popular questions from this chapter

Seeds are often treated with a fungicide for protection in poor-draining, wet environments. In a small-scale trial prior to a large-scale experiment to determine what dilution of the fungicide to apply, five treated seeds and five untreated seeds were planted in clay soil and the number of plants emerging from the treated and untreated seeds were recorded. Suppose the dilution was not effective and only four plants emerged. Let \(x\) represent the number of plants that emerged from treated seeds. a. Find the probability that \(x=4\). b. Find \(P(x \leq 3)\). c. Find \(P(2 \leq x \leq 3)\).

Let \(x\) be a binomial random variable with \(n=\) 20 and \(p=.1\) a. Calculate \(P(x \leq 4)\) using the binomial formula. b. Calculate \(P(x \leq 4)\) using Table 1 in Appendix I. c. Use the Excel output below to calculate \(P(x \leq 4)\). Compare the results of parts a, b, and c. d. Calculate the mean and standard deviation of the random variable \(x\). e. Use the results of part d to calculate the intervals \(\mu \pm \sigma, \mu \pm 2 \sigma,\) and \(\mu \pm 3 \sigma\). Find the probability that an observation will fall into each of these intervals. f. Are the results of part e consistent with Tchebysheff's Theorem? With the Empirical Rule? Why or why not?

Let \(x\) be a binomial random variable with \(n=7\), \(p=.3 .\) Find these values: a. \(P(x=4)\) b. \(P(x \leq 1)\) c. \(P(x>1)\) d. \(\mu=n p\) e. \(\sigma=\sqrt{n p q}\)

A West Coast university has found that about \(90 \%\) of its accepted applicants for enrollment in the freshman class will actually enroll. In 2012,1360 applicants were accepted to the university. Within what limits would you expect to find the size of the freshman class at this university in the fall of \(2012 ?\)

Suppose that \(10 \%\) of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in this area are randomly selected and checked for whitefly. a. What is the average number of fields sampled that are infested with whitefly? b. Within what limits would you expect to find the number of infested fields, with probability approximately \(95 \% ?\) c. What might you conclude if you found that \(x=25\) fields were infested? Is it possible that one of the characteristics of a binomial experiment is not satisfied in this experiment? Explain.

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