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Talking or texting on your cell phone can be hazardous to your health! A snapshot in USA Today reports that approximately \(23 \%\) of cell phone owners have walked into someone or something while talking on their phones. A random sample of \(n=8\) cell phone owners were asked if they had ever walked into something or someone while talking on their cell phone. The following printout shows the cumulative and individual probabilities for a binomial random variable with \(n=8\) and \(p=.23 .\) Cumulative Distribution Function Binomial with \(\mathrm{n}=8\) and \(\mathrm{p}=0.23\) $$ \begin{array}{rl} \text { X } & P(X \leq X) \\ \hline 0 & 0.12357 \\ 1 & 0.41887 \\ 2 & 0.72758 \\ 3 & 0.91201 \\ 4 & 0.98087 \\ 5 & 0.99732 \\ 6 & 0.99978 \\ 7 & 0.99999 \\ 8 & 1.00000 \end{array} $$ Probability Density Function Binomial with \(n=8\) and \(p=0.23\) $$ \begin{aligned} &\begin{array}{cc} x & P(X=x) \\ \hline 0 & 0.123574 \end{array}\\\ &\begin{array}{l} 0 & 0.123574 \\ 1 & 0.295293 \\ 2 & 0.308715 \\ 3 & 0.184427 \\ 4 & 0.068861 \\ 5 & 0.016455 \\ 6 & 0.002458 \\ 7 & 0.000210 \\ 8 & 0.000008 \end{array} \end{aligned} $$ a. Use the binomial formula to find the probability that one of the eight have walked into someone or something while talking on their cell phone. b. Confirm the results of part a using the printout. c. What is the probability that at least two of the eight have walked into someone or something while talking on their cell phone.

Short Answer

Expert verified
Answer: The probability is approximately 0.58113.

Step by step solution

01

a. Calculate the probability for one person walking into someone/something.

We need to find \(P(X=1)\) using the binomial probability formula mentioned above. $$P(X=1)=\binom{8}{1} (0.23)^{1}(1-0.23)^{8-1}$$ Now, we calculate the result: $$P(X=1)=8\times0.23\times(1-0.23)^{7}\approx 0.295293$$
02

b. Confirm the result from the printout.

To confirm our results, we look at the printout for the Probability Density Function (PDF): $$ \begin{array}{cc} x & P(X=x) \\ \hline 1 & 0.295293 \end{array} $$ Our calculated probability matches the given printout: \(0.295293\). Hence, we have confirmed the result for part a.
03

c. Calculate the probability for at least two people walking into someone/something.

To find the probability of at least two people walking into someone or something, we will first calculate the probability of less than two people walking into someone or something and then subtract it from 1. Using the printout of the Cumulative Distribution Function (CDF): $$P(X\leq1)=0.41887$$ Now, we will calculate the complimentary event \(P(X\geq 2)\) by subtracting the cumulative probability of less than or equal to 1 as follows: $$P(X\geq2)=1-P(X\leq1)=1-0.41887 \approx 0.58113$$ The probability that at least two out of the eight cell phone owners have walked into someone or something while talking on their cell phone is approximately \(0.58113\).

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Key Concepts

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

Cumulative Distribution Function
Understanding the cumulative distribution function (CDF) is crucial when dealing with random events. The CDF for a random variable X represents the probability that X will take a value less than or equal to a certain value x. In other words, it gives you the cumulative probability of the random variable from the lowest possible outcome up to that point.

For example, in the context of the cell phone owners' problem mentioned in the exercise, the CDF tells us the probability that up to a certain number of people have walked into someone or something while using their phone. If you see a value of 0.72758 for X = 2 in the CDF, it means there's a 72.758% chance that two or fewer cell phone owners in the sample have had such an accident.

To calculate the CDF formally, you sum the probabilities of the event happening for all values up to the given point. For binomial distributions, you can use a formula, calculator, or table to find CDF values. Remember, understanding the CDF is vital for finding probabilities for 'at least' or 'less than' scenarios like part c of our problem.
Probability Density Function
While the cumulative distribution function provides cumulative probabilities, the probability density function (PDF) deals with specific probabilities. In the scenario of binomial probability, such as the cell phone users' predicament, the PDF gives us the probability of observing a precise number of occurrences of an event.

For each possible outcome x, the PDF details the likelihood, represented as P(X=x). This is paramount when you want to know the exact probability of a certain number of events occurring, such as exactly one person walking into something while on the phone, which was calculated as a probability of about 29.5293% in the exercise.

The binomial formula is typically used to calculate the probability for each outcome in a discrete setting like this. It's important to grasp the PDF, especially when you require the likelihood of a single, exact event rather than a range of outcomes.
Random Variable
A random variable is a foundational concept in probability and statistics that quantifies the outcomes of a random phenomenon. In simpler terms, it's a variable whose values depend on the outcomes of a random process. There are two types of random variables: discrete and continuous. Binomial experiments, like the one involving the cell phone users, involve discrete random variables because the outcomes can only take on a countable number of values.

In our case, the random variable X represents the number of cell phone owners who walk into someone or something while talking on their phones out of a sample size of eight. Each value that X can take has an associated probability, which can be understood using the PDF and is summed up in the CDF. Random variables allow us to formally work with randomness and to assign probabilities to different outcomes, making them indispensable in probabilistic modeling and statistical inference.

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Most popular questions from this chapter

Use the formula for the binomial probability distribution to calculate the values of \(p(x)\) and construct the probability histogram for \(x\) when \(n=6\) and \(p=.2\). [HINT: Calculate \(P(x=k\) ) for seven different values of \(k\).

Let \(x\) be the number of successes observed in a sample of \(n=5\) items selected from a population of \(N=10 .\) Suppose that of the \(N=10\) items, \(M=6\) are considered "successes." Find the probabilities in Exercises \(11-13 .\) The probability of observing exactly two successes.

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 following Excel output 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? Excel output for Exercise 33: Binomial with \(n=20\) and \(p=.1\) $$ \begin{array}{|c|c|c|c|c|} \hline & \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D} \\ \hline 1 & \mathrm{x} & \mathrm{p}(\mathrm{x}) & \mathrm{x} & \mathrm{p}(\mathrm{x}) \\ \hline 2 & 0 & 0.1216 & 11 & 7 \mathrm{E}-07 \\ \hline 3 & 1 & 0.2702 & 12 & 5 \mathrm{E}-08 \\ \hline 4 & 2 & 0.2852 & 13 & 4 \mathrm{E}-09 \\ \hline 5 & 3 & 0.1901 & 14 & 2 \mathrm{E}-10 \\ \hline 6 & 4 & 0.0898 & 15 & 9 \mathrm{E}-12 \\ \hline 7 & 5 & 0.0319 & 16 & 3 \mathrm{E}-13 \\ \hline 8 & 6 & 0.0089 & 17 & 8 \mathrm{E}-15 \\ \hline 9 & 7 & 0.0020 & 18 & 2 \mathrm{E}-16 \\ \hline 10 & 8 & 0.0004 & 19 & 2 \mathrm{E}-18 \\ \hline 11 & 9 & 0.0001 & 20 & 1 \mathrm{E}-20 \\ \hline 12 & 10 & 0.0000 & & \\ \hline \end{array} $$

If \(x\) has a binomial distribution with \(p=.5\), will the shape of the probability distribution be symmetric, skewed to the left, or skewed to the right?

A fire-detection device uses three temperature-sensitive cells acting independently of one another so that any one or more can activate the alarm. Each cell has a probability \(p=.8\) of activating the alarm when the temperature reaches \(57^{\circ} \mathrm{C}\) or higher. Let \(x\) equal the number of cells activating the alarm when the temperature reaches \(57^{\circ} \mathrm{C}\). a. Find the probability distribution of \(x\). b. Find the probability that the alarm will function when the temperature reaches \(57^{\circ} \mathrm{C}\). c. Find the expected value and the variance for the random variable \(x\).

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