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Are your finances, buying habits, medical records, and phone calls really private? A real concern for many adults is that computers and the Internet are reducing privacy. A survey conducted by Peter D. Hart Research Associates for the Shell Poll was reported in USA Today. According to the survey, \(37 \%\) of adults are concerned that employers are monitoring phone calls. Use the binomial distribution formula to calculate the probability that (a) out of five adults, none is concerned that employers are monitoring phone calls. (b) out of five adults, all are concerned that employers are monitoring phone calls. (c) out of five adults, exactly three are concerned that employers are monitoring phone calls.

Short Answer

Expert verified
(a) 0.069, (b) 0.005, (c) 0.215.

Step by step solution

01

Understand the Binomial Distribution

The binomial distribution describes the number of successes in a fixed number of independent Bernoulli trials. Success is defined as an adult being concerned, with probability of success given by \( p = 0.37 \). For our problem, the number of trials \( n = 5 \).
02

Recognize the Binomial Distribution Formula

The binomial probability formula is given by: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \( \binom{n}{k} \) is the binomial coefficient, \( p \) is the probability of success, and \( k \) is the number of successes.
03

Calculate Probability for No Concern

For part (a), we calculate the probability that none out of five adults (\( k = 0 \)) are concerned: \[ P(X=0) = \binom{5}{0} \times (0.37)^0 \times (0.63)^5 = 1 \times 1 \times (0.63)^5 = (0.63)^5 \approx 0.069. \]
04

Calculate Probability for All Concern

For part (b), we calculate the probability that all five adults (\( k = 5 \)) are concerned: \[ P(X=5) = \binom{5}{5} \times (0.37)^5 \times (0.63)^0 = 1 \times (0.37)^5 \times 1 = (0.37)^5 \approx 0.005. \]
05

Calculate Probability for Exactly Three Concern

For part (c), we calculate the probability that exactly three out of five adults (\( k = 3 \)) are concerned: \[ P(X=3) = \binom{5}{3} \times (0.37)^3 \times (0.63)^2 = 10 \times (0.37)^3 \times (0.63)^2 \approx 0.215. \]

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Key Concepts

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

Probability
Probability is all about predicting the likelihood of different outcomes. It's like trying to guess the chances of rain tomorrow or how likely you are to roll a six on a dice. In math, probability is expressed as a number between 0 and 1, where 0 means there's no chance at all, and 1 means it's definitely going to happen.
When it comes to daily concerns, like the scenario where 37% of adults are worried about privacy, probability helps us understand the chances of these events happening in a group. This is where the concept of binomial distribution comes in handy, as it allows us to calculate probabilities for events, like some adults being concerned about monitored phone calls.
By using probability, we can determine how often we can expect a certain outcome, such as none, a few, or all adults being concerned. This helps in decision-making and understanding risks. Knowing probabilities can change how we interpret situations.
Bernoulli Trials
Bernoulli Trials are the building blocks of the binomial distribution. Imagine flipping a coin—each flip is a Bernoulli Trial. It only has two possible outcomes: heads or tails, just like a Bernoulli Trial can only have success or failure.
In our scenario, each adult's reaction (either concerned or not) is a Bernoulli trial. With a probability of 0.37 for a concerned adult (success), each individual's response is a separate event.
Bernoulli Trials are unique because each event is independent, meaning that one adult being concerned does not affect another's concern. This independence is crucial because it assures that each trial is statistically fair, allowing us to predict the group behavior accurately.
Overall, Bernoulli Trials help break down complex probabilistic scenarios into simpler, manageable parts.
Binomial Coefficient
The Binomial Coefficient is a fancy term that shows up when you need to calculate how many ways you can arrange a group. It’s often represented in math as \( \binom{n}{k} \), read as "n choose k". This number tells us how many ways \( k \) successes can occur out of \( n \) trials.
In our case of concerned adults out of five, the binomial coefficient helps calculate how these concerns can be arranged across a group.
For example, to find how exactly 3 adults out of 5 might be concerned, the \( \binom{5}{3} \) term calculates how many different combinations could lead to exactly three adults being concerned.
Using binomial coefficients, we can efficiently determine combinations, which is essential for calculating the probabilities in a binomial distribution scenario.
Independent Events
Independent events are events where the outcome of one does not affect the other. Think of rolling two dice. The result of the first roll does not change what happens with the second roll.
In our binomial distribution situation, each adult’s concern is an independent event. Whether one adult is concerned does not influence another adult's attitude.
This independence is crucial in calculating probabilities using the binomial formula because it ensures that the probability of one event stays constant no matter what happens with other events.
By treating each adult's concern as independent, it simplifies our calculations and makes sure that probabilities are calculated correctly, leading to precise predictions about group behaviors.

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Most popular questions from this chapter

Let \(r\) be a binomial random variable representing the number of successes out of \(n\) trials. (a) Explain why the sample space for \(r\) consists of the set \(\\{0,1,2, \ldots, n\\}\) and why the sum of the probabilities of all the entries in the entire sample space must be 1. (b) Explain why \(P(r \geq 1)=1-P(0)\) (c) Explain why \(P(r \geq 2)=1-P(0)-P(1)\) (d) Explain why \(P(r \geq m)=1-P(0)-P(1)-\cdots-P(m-1)\) for \(1 \leq m \leq n\).

Golf Norb and Gary are entered in a local golf tournament. Both have played the local course many times. Their scores are random variables with the following means and standard deviations. $$\text { Norb, } x_{1}: \mu_{1}=115 ; \sigma_{1}=12 \quad \text { Gary, } x_{2}: \mu_{2}=100 ; \sigma_{2}=8$$ In the tournament, Norb and Gary are not playing together, and we will assume their scores vary independently of each other. (a) The difference between their scores is \(W=x_{1}-x_{2} .\) Compute the mean, variance, and standard deviation for the random variable \(W.\) (b) The average of their scores is \(W=0.5 x_{1}+0.5 x_{2}\). Compute the mean, variance, and standard deviation for the random variable \(W.\) (c) The tournament rules have a special handicap system for each player. For Norb, the handicap formula is \(L=0.8 x_{1}-2 .\) Compute the mean, variance, and standard deviation for the random variable \(L.\) (d) For Gary, the handicap formula is \(L=0.95 x_{2}-5 .\) Compute the mean, variance, and standard deviation for the random variable \(L.\)

The college hiking club is having a fundraiser to buy new equipment for fall and winter outings. The club is selling Chinese fortune cookies at a price of 1 dollar per cookie. Each cookie contains a piece of paper with a different number written on it. A random drawing will determine which number is the winner of a dinner for two at a local Chinese restaurant. The dinner is valued at 35 dollar. since the fortune cookies were donated to the club, we can ignore the cost of the cookies. The club sold 719 cookies before the drawing. (a) Lisa bought 15 cookies. What is the probability she will win the dinner for two? What is the probability she will not win? (b) Lisa's expected earnings can be found by multiplying the value of the dinner by the probability that she will win. What are Lisa's expected earnings? How much did she effectively contribute to the hiking club?

USA Today reports that about \(25 \%\) of all prison parolees become repeat offenders. Alice is a social worker whose job is to counsel people on parole. Let us say success means a person does not become a repeat offender. Alice has been given a group of four parolees. (a) Find the probability \(P(r)\) of \(r\) successes ranging from 0 to 4 (b) Make a histogram for the probability distribution of part (a). (c) What is the expected number of parolees in Alice's group who will not be repeat offenders? What is the standard deviation? (d) How large a group should Alice counsel to be about \(98 \%\) sure that three or more parolees will not become repeat offenders?

The one-time fling! Have you ever purchased an article of clothing (dress, sports jacket, etc.), worn the item once to a party, and then returned the purchase? This is called a one-time fling. About \(10 \%\) of all adults deliberately do a one-time fling and feel no guilt about it (Source: Are You Normal?, by Bernice Kanner, St. Martin's Press). In a group of seven adult friends, what is the probability that (a) no one has done a one-time fling? (b) at least one person has done a one-time fling? (c) no more than two people have done a one-time fling?

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