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A particular telephone number is used to receive both voice calls and fax messages. Suppose that \(25 \%\) of the incoming calls involve fax messages, and consider a sample of 25 incoming calls. What is the probability that a. At most 6 of the calls involve a fax message? b. Exactly 6 of the calls involve a fax message? c. At least 6 of the calls involve a fax message? d. More than 6 of the calls involve a fax message?

Short Answer

Expert verified
For (a) \(P(X \leq 6)\), (b) \(P(X = 6)\), (c) \(P(X \geq 6)\), and (d) \(P(X > 6)\), calculate using the binomial distribution.

Step by step solution

01

Define the Distribution

The problem can be modeled using the binomial distribution, where the number of trials \(n\) is 25 (the number of incoming calls), and the probability of success \(p\) is 0.25 (the probability a call is a fax). Each call is either a fax or not, making it a binomial scenario.
02

Binomial Formula

The probability of exactly \(k\) fax calls out of \(n\) incoming calls is given by the formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \(\binom{n}{k}\) is the binomial coefficient.
03

Calculate part (a)

To find the probability that at most 6 calls are fax, calculate: \( P(X \leq 6) = P(X=0) + P(X=1) + \ldots + P(X=6) \). This involves summing up the probabilities from \(P(X=0)\) to \(P(X=6)\).
04

Calculate part (b)

To find the probability of exactly 6 fax calls, calculate: \( P(X = 6) = \binom{25}{6} (0.25)^6 (0.75)^{19} \). Compute the binomial coefficients and the powers to get this probability.
05

Calculate part (c)

To find the probability that at least 6 calls involve fax messages, calculate: \( P(X \geq 6) = 1 - P(X \leq 5) \). Use the cumulative probability for \(P(X \leq 5)\) and subtract from 1.
06

Calculate part (d)

To find the probability that more than 6 calls are fax, calculate: \( P(X > 6) = 1 - P(X \leq 6) \). This involves subtracting the probability of at most 6 fax calls from 1.

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Key Concepts

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

Probability Calculation
Probability calculation is a fundamental concept in statistics used to determine the likelihood of various outcomes. In the context of a binomial distribution, this involves calculating the probability of a specific number of events, such as receiving fax calls among a set of incoming calls. The key to probability calculation is understanding how to apply the binomial formula effectively.
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
This formula helps us determine the probability of exactly \(k\) successes (fax calls, in this case) in \(n\) trials (the total calls). Here's a quick walkthrough:
  • \(\binom{n}{k}\) represents the binomial coefficient, which means the number of ways to choose \(k\) successes from \(n\) trials.
  • \(p^k\) stands for the probability of getting \(k\) fax calls, raised to the number of successful calls.
  • \((1-p)^{n-k}\) reflects the likelihood of the remaining calls not being faxes.
Consider the probability of exactly 6 fax messages from 25 calls, using the above formula. This is a direct calculation of the probability of an exact outcome, involving the multiplication of probabilities for both events: fax and no fax.
Binomial Coefficient
The binomial coefficient is a significant component of the binomial distribution equation. It is denoted as \(\binom{n}{k}\), which can be read as "n choose k". This part of the formula counts how many distinct ways you can choose \(k\) successes from \(n\) trials.
To compute \(\binom{n}{k}\), use the formula:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Here's a breakdown:
  • \(n!\) is the factorial of \(n\), meaning the product of all numbers from 1 up to \(n\).
  • \(k!\) is the factorial of \(k\), and \((n-k)!\) is the factorial of \((n-k)\).
In our exercise scenario, if you are calculating \(\binom{25}{6}\), this computation would involve quite a few multiplications and divisions, but it provides the number of possible ways to obtain 6 fax calls from 25 calls. Knowing how to accurately calculate the binomial coefficient is crucial in determining the probability of specific outcomes in a binomial distribution.
Cumulative Probability
Cumulative probability is an extension of individual probability calculations, used to find the probability of a set of outcomes grouped together. It answers questions like "at most", "at least", or "less than" certain events happening.
To find the probability that at most 6 calls are fax messages among 25:
\[ P(X \leq 6) = P(X=0) + P(X=1) + \ldots + P(X=6) \]
This method involves computing and summing the probabilities from 0 up to 6 fax calls. It's useful for cases where multiple outcomes are possible, and we need the total likelihood of those grouped events occurring.
For "at least" scenarios, use the complementary probability:
\[ P(X \geq 6) = 1 - P(X \leq 5) \]
This involves subtracting the cumulative probability up to 5 from 1 to find the likelihood of having 6 or more successes (fax calls). Understanding cumulative probability helps deal with ranges of desired outcomes rather than single counts.
Random Variable
In probability theory, a random variable is a variable whose values depend on outcomes of a random phenomenon. In our discussion on the binomial distribution, the random variable \(X\) represents the number of fax calls received.
Key characteristics of a random variable in binomial distribution:
  • It can take integer values ranging from 0 to \(n\), where \(n\) is the number of trials or events.
  • The probability associated with \(X\) is determined using the binomial formula.
  • The random variable in this exercise is discrete, not continuous, since it represents countable outcomes.
For example, when we talk about calculating \(P(X=6)\), we are finding the probability that the random variable \(X\) (number of fax calls) equals 6 out of the total number of events (25 calls). Understanding the concept of a random variable is fundamental in linking real-world events, like receiving calls, to mathematical probabilities.

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Most popular questions from this chapter

An instructor who taught two sections of engineering statistics last term, the first with 20 students and the second with 30 , decided to assign a term project. After all projects had been turned in, the instructor randomly ordered them before grading. Consider the first 15 graded projects. a. What is the probability that exactly 10 of these are from the second section? b. What is the probability that at least 10 of these are from the second section? c. What is the probability that at least 10 of these are from the same section? d. What are the mean value and standard deviation of the number among these 15 that are from the second section? e. What are the mean value and standard deviation of the number of projects not among these first 15 that are from the second section?

Let \(X\) have a Poisson distribution with parameter \(\mu\). Show that \(E(X)=\mu\) directly from the definition of expected value. [Hint: The first term in the sum equals 0 , and then \(x\) can be canceled. Now factor out \(\mu\) and show that what is left sums to 1.]

Give three examples of Bernoulli rv's (other than those in the text).

Let \(X=\) the outcome when a fair die is rolled once. If before the die is rolled you are offered either \((1 / 3.5)\) dollars or \(h(X)=1 / X\) dollars, would you accept the guaranteed amount or would you gamble? [Note: It is not generally true that \(1 / E(X)=E(1 / X)\).]

If the sample space \(S\) is an infinite set, does this necessarily imply that any rv \(X\) defined from \(\&\) will have an infinite set of possible values? If yes, say why. If no, give an example.

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