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Chapter 5: Problem 31

A Randstad/Harris interactive survey reported that \(25 \%\) of employees said their company is loyal to them (USA Today, November 11,2009 ). Suppose 10 employees are selected randomly and will be interviewed about company loyalty. a. Is the selection of 10 employees a binomial experiment? Explain. b. What is the probability that none of the 10 employees will say their company is loyal to them? c. What is the probability that 4 of the 10 employees will say their company is loyal to them? d. What is the probability that at least 2 of the 10 employees will say their company is loyal to them?

Short Answer

Expert verified
a. Yes, it is a binomial experiment. b. Approximately 0.0563. c. Approximately 0.145998. d. Approximately 0.756.

Step by step solution

01

Understanding a Binomial Experiment

A binomial experiment has four characteristics: fixed number of trials, each trial is independent, only two possible outcomes (success or failure), and the probability of success is constant. In this context, selecting 10 employees means we have a fixed number of trials (10), assuming each employee's opinion is independent of the others. The outcome for each trial can be categorized as a success (the employee says the company is loyal) or failure (the employee does not), with a constant probability of success (25%) for each trial.
02

Defining parameters for the binomial probability

Since the selection of 10 employees meets the criteria for a binomial experiment: "Having a fixed number of trials (10), independent trials, two outcomes, and a constant probability," this is indeed a binomial experiment. We define the success probability as\(p = 0.25\) and the number of trials \(n = 10\).
03

Calculation of probability for zero successes

To find the probability that none of the 10 employees say their company is loyal, use the binomial probability formula:\[P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\]Set \(k = 0\), \(n = 10\), and \(p = 0.25\):\[P(X=0) = \binom{10}{0} (0.25)^0 (0.75)^{10} = 1 \times 1 \times (0.75)^{10}\]Calculating gives approximately \(P(X=0) \approx 0.0563\).
04

Calculation of probability for exactly 4 successes

For the probability that exactly 4 employees out of 10 say their company is loyal, use the same formula with \(k=4\), \(n=10\), and \(p=0.25\):\[P(X=4) = \binom{10}{4} (0.25)^4 (0.75)^6\]Calculating gives \(P(X=4) \approx 0.145998\).
05

Calculation of probability for at least 2 successes

To find the probability of at least 2 employees saying their company is loyal, calculate 1 minus the probability of having less than 2 successes (0 or 1 success):\[P(X \geq 2) = 1 - (P(X=0) + P(X=1))\]We've already calculated \(P(X=0)\). Now, calculate \(P(X=1)\):\[P(X=1) = \binom{10}{1} (0.25)^1 (0.75)^9\]Calculate \(P(X=1) \approx 0.1877\). Then:\[P(X \geq 2) = 1 - (0.0563 + 0.1877) = 1 - 0.244\]So, \(P(X \geq 2) \approx 0.756\).

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Key Concepts

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

Probability Calculation
Probability calculation in binomial experiments is straightforward once you know the right formula and how to use it. Each probability in a binomial setting can be calculated using the binomial probability formula, which tells us the likelihood of a certain number of successes in a given number of trials.

Here's how it works:
  • First, you need the number of trials: this is the total number of times you are conducting the experiment. In our case, it's 10 employees.
  • Next, decide on the number of successes you're interested in. This could be any number from zero to the total number of trials.
  • You'll also need the probability of success for a single trial, which remains constant across all trials.
Once you have these, you can plug them into the binomial probability formula: \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]This formula calculates the probability of getting exactly "k" successes in "n" trials, with "p" being the probability of success in each trial.
Binomial Probability Formula
The binomial probability formula is central to solving binomial experiments. Understanding this formula will help you swiftly solve many binomial probability questions. It calculates the probability of achieving a certain number of successes in a set number of independent trials with a constant probability of success across each trial.

The formula itself is: \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \] where:
  • \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose "k" successes from "n" trials.
  • \( p^k \) is the probability of having "k" successes.
  • \((1-p)^{n-k}\) is the probability of the rest of the trials being failures.
This formula gives you a clear, mathematical way to determine the probability of any number of successes in your trials, which is incredibly useful in statistics.
Independent Trials
In a binomial experiment, each trial is independent, which is a crucial aspect to consider. **Independent trials** mean that the outcome of one trial does not affect or rely on the outcome of any other trial.

This independence ensures that each trial remains unbiased and maintains a constant probability of success.
For example, when we conduct a survey and ask 10 employees about their perception of company loyalty, each employee's response is considered independent of others.

Key attributes of independent trials:
  • Each trial must not influence any other.
  • The probability of success must remain the same for every trial.
  • There are only two outcomes in each trial: success or failure.
This independence is what makes a sequence of trials fit neatly into the framework of a binomial experiment, allowing for straightforward probability calculations using the binomial formula.

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Most popular questions from this chapter

Consider a binomial experiment with \(n=10\) and \(p=.10\) a. Compute \(f(0)\) b. Compute \(f(2)\) c. Compute \(P(x \leq 2)\) d. Compute \(P(x \geq 1)\) e. Compute \(E(x)\) f. Compute \(\operatorname{Var}(x)\) and \(\sigma\).

The National Basketball Association (NBA) records a variety of statistics for each team. Two of these statistics are the percentage of field goals made by the team and the percentage of three-point shots made by the team. For a portion of the 2004 season, the shooting records of the 29 teams in the NBA showed that the probability of scoring two points by making a field goal was \(.44,\) and the probability of scoring three points by making a threepoint shot was .34 (NBA website, January 3,2004 ). a. What is the expected value of a two-point shot for these teams? b. What is the expected value of a three-point shot for these teams? c. If the probability of making a two-point shot is greater than the probability of making a three-point shot, why do coaches allow some players to shoot the three-point shot if they have the opportunity? Use expected value to explain your answer.

A psychologist determined that the number of sessions required to obtain the trust of a new patient is either \(1,2,\) or \(3 .\) Let \(x\) be a random variable indicating the number of sessions required to gain the patient's trust. The following probability function has been proposed. \\[ f(x)=\frac{x}{6} \quad \text { for } x=1,2, \text { or } 3 \\] a. Is this probability function valid? Explain. b. What is the probability that it takes exactly two sessions to gain the patient's trust? c. What is the probability that it takes at least two sessions to gain the patient's trust?

According to a survey conducted by TD Ameritrade, one out of four investors have exchange-traded funds in their portfolios (USA Today, January 11,2007 ). Consider a sample of 20 investors. a. Compute the probability that exactly four investors have exchange-traded funds in their portfolios. b. Compute the probability that at least two of the investors have exchange- traded funds in their portfolios. c. If you found that exactly 12 of the investors have exchange-traded funds in their portfolios, would you doubt the accuracy of the survey results? d. Compute the expected number of investors who have exchange-traded funds in their portfolios.

Phone calls arrive at the rate of 48 per hour at the reservation desk for Regional Airways. a. Compute the probability of receiving three calls in a 5 -minute interval of time. b. Compute the probability of receiving exactly 10 calls in 15 minutes. c. Suppose no calls are currently on hold. If the agent takes 5 minutes to complete the current call, how many callers do you expect to be waiting by that time? What is the probability that none will be waiting? d. If no calls are currently being processed, what is the probability that the agent can take 3 minutes for personal time without being interrupted by a call?
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