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Chapter 5: Problem 22

Assume that when adults with smartphones are randomly selected, \(54 \%\) use them in meetings or classes (based on data from an LG Smartphone survey). If 20 adult smartphone users are randomly selected, find the probability that exactly 15 of them use their smartphones in meetings or classes.

Short Answer

Expert verified
The probability is 0.1285.

Step by step solution

01

- Understand the problem

We need to find the probability that exactly 15 out of 20 selected adults use their smartphones in meetings or classes, given that 54% of adults use them in such scenarios.
02

- Identify the type of problem

This is a binomial probability problem where we need to find the probability of a certain number of successes (people using smartphones in meetings) in a fixed number of trials (total people selected).
03

- Define variables for the binomial probability

Here: - The number of trials, n = 20- The number of successes, k = 15- The probability of success on a single trial, p = 0.54
04

- Write the binomial probability formula

The formula for binomial probability is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \(\binom{n}{k}\) is the binomial coefficient.
05

- Calculate the binomial coefficient

First, compute \(\binom{n}{k}\) where \(\binom{n}{k} = \frac{n!}{k!(n-k)!} \):\( \binom{20}{15} = \frac{20!}{15!5!} = 15504 \).
06

- Compute the probabilities

Calculate \( p^k \) and \( (1-p)^{n-k} \):\( p^k = 0.54^{15} \) and \( (1-p)^{n-k} = (0.46)^5 \).
07

- Combine into the formula

Multiply the binomial coefficient by the probabilities: \[ P(X = 15) = 15504 \times 0.54^{15} \times 0.46^5 \].
08

- Calculate the final probability

Perform the calculations: \[ 0.54^{15} \approx 3.717 \times 10^{-4} \] and \[ 0.46^5 \approx 0.0453 \]. Multiply all values together: \[ 15504 \times 3.717 \times 10^{-4} \times 0.0453 \approx 0.1285 \].

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Key Concepts

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

binomial distribution
To apply the binomial formula, we need to identify our n (number of trials), k (number of successes), and p (probability of success in one trial). These values are then substituted into the formula for the final calculation.
probability calculation
By following these detailed steps, one can calculate the binomial probability for any similar situation.
binomial coefficient
This coefficient indicates there are 15,504 different ways to arrange 15 successes and 5 failures in twenty trials. Understanding how to compute and interpret this value is key to solving binomial probability problems.
statistics problems
  • Market research: Predicting consumer behavior, like how many people will use a smartphone in specific situations.
  • Medical studies: Determining the probability of patients responding to a treatment.
  • Quality control: Assessing the chance of defects in a batch of products.

In these examples, we can set up a similar structure: specify n (total trials), p (probability of success per trial), and k (desired number of successes) to calculate probabilities using the binomial formula.
Whether predicting the number of successful outcomes in trials or informing decisions based on probability, mastering binomial distributions and calculations is an invaluable statistical skill.

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Most popular questions from this chapter

Assume that the Poisson distribution applies; assume that the mean number of Atlantic hurricanes in the United States is 6.1 per year, as in Example \(I\); and proceed to find the indicated probability. Hurricanes a. Find the probability that in a year, there will be 5 hurricanes. b. In a 55 -year period, how many years are expected to have 5 hurricanes? c. How does the result from part (b) compare to the recent period of 55 years in which 8 years had 5 hurricanes? Does the Poisson distribution work well here?

Use the Poisson distribution to find the indicated probabilities. Kicks A classical example of the Poisson distribution involves the number of deaths caused by horse kicks to men in the Prussian Army between 1875 and 1894 Data for 14 corps were combined for the 20 -year period, and the 280 corps- years included a total of 196 deaths. After finding the mean number of deaths per corps-year, find the probability that a randomly selected corps-year has the following numbers of deaths: (a) \(0,\) (b) \(1,\) (c) 2 , (d) \(3,(\mathrm{e})\) 4. The actual results consisted of these frequencies: 0 deaths (in 144 corps-years); 1 death (in 91 corps-years); 2 deaths (in 32 corps-ycars); 3 deaths (in 11 corps- years); 4 deaths (in 2 corps-years). Compare the actual results to those expected by using the Poisson probabilities. Does the Poisson distribution serve as a good tool for predicting the actual results?

Use the Poisson distribution to find the indicated probabilities. In Exercise 1 "Notation" we noted that in analyzing hits by V-1 buzz bombs in World War II, South London was partitioned into 576 regions, each with an area of \(0.25 \mathrm{km}^{2} .\) A total of 535 bombs hit the combined area of 576 regions. a. Find the probability that a randomly selected region had exactly 2 hits. b. Among the 576 regions, find the expected number of regions with exactly 2 hits. c. How does the result from part (b) compare to this actual result: There were 93 regions that had exactly 2 hits?

Use the Poisson distribution to find the indicated probabilities. In a recent year, NYU-Langone Medical Center had 4221 birhs. Find the mean number of births per day, then use that result to find the probability that in a day, there are 15 births. Does it appear likely that on any given day, there will be exactly 15 births?

There is a \(1 / 292,201,338\) probability of winning the Powerball lottery jackpot with a single ticket. Assume that you purchase a ticket in each of the next 5200 different Powerball games that are run over the next 50 years. Find the probability of winning the jackpot with at least one of those tickets. Is there a good chance that you would win the jackpot at least once in 50 years?
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