/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 A business has six customer serv... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A business has six customer service telephone lines. Let \(x\) denote the number of lines in use at any given time. Suppose that the probability distribution of \(x\) is as follows: $$ \begin{array}{lccccccc} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ p(x) & 0.10 & 0.15 & 0.20 & 0.25 & 0.20 & 0.06 & 0.04 \end{array} $$ Write each of the following events in terms of \(x,\) and then calculate the probability of each one: a. At most three lines are in use b. Fewer than three lines are in use c. At least three lines are in use d. Between two and five lines (inclusive) are in use e. Between two and four lines (inclusive) are not in use f. At least four lines are not in use

Short Answer

Expert verified
The probabilities of the events are as follows: \n a. 0.70, \n b. 0.45, \n c. 0.55, \n d. 0.71, \n e. 0.45, \n f. 0.45 \n These are calculated by adding up the corresponding probabilities from the given probability distribution.

Step by step solution

01

Translate Statements

First, each event needs to be translated into terms of \(x\). In each statement, the terms 'at most', 'fewer than', 'at least' and 'between...inclusive' refers to ranges of x-values. Thus, these events corresponds to the following x-values: a. 'At most three lines are in use' implies \(x = 0, 1, 2, 3\). b. 'Fewer than three lines are in use' implies \(x = 0, 1, 2\).c. 'At least three lines are in use' implies \(x = 3, 4, 5, 6\).d. 'Between two and five lines are in use' implies \(x = 2, 3, 4, 5\).e. 'Between two and four lines are not in use' implies \(x = 2, 3, 4\).f. 'At least four lines are not in use' implies \(x = 0, 1, 2\).
02

Calculate Probabilities

Now, to calculate the probability of each event occurring, the associated probabilities from given distribution should be added up for all corresponding x-values. Here are the probability calculations for each event using given distribution:a. \(P(x \leq 3) = P(x=0) + P(x=1) + P(x=2) + P(x=3) = 0.10 + 0.15 + 0.20 + 0.25 = 0.70\).b. \(P(x < 3) = P(x=0) + P(x=1) + P(x=2) = 0.10 + 0.15 + 0.20 = 0.45\).c. \(P(x \geq 3) = P(x=3) + P(x=4) + P(x=5) + P(x=6) = 0.25 + 0.20 + 0.06 + 0.04 = 0.55\).d. \(P(2 \leq x \leq 5) = P(x=2) + P(x=3) + P(x=4) + P(x=5) = 0.20 + 0.25 + 0.20 + 0.06 = 0.71\).e. \(P(2 \leq x \leq 4) = P(x=0) + P(x=1) + P(x=2) = 0.10 + 0.15 + 0.20 = 0.45\).f. \(P(x \leq 2) = P(x=0) + P(x=1) + P(x=2) = 0.10 + 0.15 + 0.20 = 0.45\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Event Probability
Understanding event probability is crucial when working with probability distributions. An event is any set of outcomes we are interested in, within a probability space. In our exercise, an event like "at most three lines are in use" involves specific outcomes from the random variable.
  • "At most" means considering all values up to and including the designated number. For example, "at most three" means we consider x=0, x=1, x=2, and x=3.
  • "Fewer than three" involves all values less than three, which would be x=0, x=1, and x=2.
  • "At least three" means verifing all outcomes from three upwards, including x=3, x=4, x=5, and x=6.
Calculating the probability of these events involves summing the probabilities of these specified outcomes. For instance, using the data provided, the probability that at most three lines are in use is 0.70, by adding the probabilities of x-values 0 through 3. Calculating event probabilities helps to predict and understand potential outcomes in uncertain scenarios.
Random Variable
A random variable is a variable that takes on values determined by the outcome of a random phenomenon. In simple terms, it's how we represent outcomes in mathematical terms. In our exercise, the random variable is denoted by \(x\), representing the number of telephone lines in use.Random variables can be of different types: discrete random variables, which take on a countable number of distinct values, and continuous random variables, which can take on any value in a continuum.
  • Our variable \(x\) is discrete, as it can only assume specific whole number values from 0 to 6, based on how many lines are in use.
  • The actual value \(x\) takes depends on the outcome of the random process; here, how many customers are calling in at the same time.
By using random variables like \(x\), we're able to model and evaluate different scenarios involving uncertainty, like how many telephone lines are likely to be in use at any point in time. The accompanying probability distribution gives the likelihood of each possible number of lines being used.
Discrete Probability
Discrete probability is the branch of probability that deals with discrete random variables. Discrete random variables, like our \(x\) in the exercise, can only take specific, countable values. This makes calculating probabilities clearer, as we can list all possible outcomes.For each value that our random variable takes, a probability is assigned, which must meet these criteria:
  • The probability of any specific outcome is between 0 and 1.
  • The sum of all probabilities must equal 1, ensuring that they cover the entire sample space.
  • For example, the probabilities for \(x=0,1,2,3,4,5,6\) add up to 1, confirming that all potential numbers of lines in use are accounted for.
This makes methods like summing probabilities straightforward for determining the likelihood of various events. Events are composed of multiple specific outcomes, and their probabilities reflect how likely each event is within the discrete framework. Discrete probability distributions thus serve as a powerful tool for predicting outcomes in environments with finite possibilities, aiding in decision-making and planning where exact numbers are involved.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. The probability distribution of \(y\) is as follows: $$ \begin{array}{lccccc} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & 0.01 \end{array} $$ a. Calculate and interpret \(\mu_{y}\). (Hint: See Example 6.13) b. In the long run, for what percentage of cartons is the number of broken eggs less than \(\mu_{y} ?\) Does this surprise you? c. Explain why \(\mu_{y}\) is not equal to \(\frac{0+1+2+3+4}{5}=2.0\).

Refer to the previous exercise. Suppose that there are two machines available for cutting corks. The machine described in the preceding problem produces corks with diameters that are approximately normally distributed with mean \(3 \mathrm{~cm}\) and standard deviation \(0.1 \mathrm{~cm}\). The second machine produces corks with diameters that are approximately normally distributed with mean \(3.05 \mathrm{~cm}\) and standard deviation \(0.01 \mathrm{~cm}\). Which machine would you recommend? (Hint: Which machine would produce fewer defective corks?)

Let \(z\) denote a random variable having a normal distribution with \(\mu=0\) and \(\sigma=1 .\) Determine each of the following probabilities: a. \(P(z < 0.10)\) b. \(P(z < -0.10)\) c. \(P(0.40 < z < 0.85)\) d. \(P(-0.85 < z < -0.40)\) e. \(P(-0.40 < z < 0.85)\) f. \(P(z > \- 1.25)\) g. \(P(z < -1.50\) or \(z > 2.50)\)

Suppose that the distribution of typing speed in words per minute (wpm) for experienced typists using a new type of split keyboard can be approximated by a normal curve with mean 60 wpm and standard deviation 15 wpm ("The Effects of Split Keyboard Geometry on Upper Body Postures, Ergonomics [2009]: 104-111). a. What is the probability that a randomly selected typist's speed is at most 60 wpm? Less than 60 wpm? b. What is the probability that a randomly selected typist's speed is between 45 and 90 wpm? c. Would you be surprised to find a typist in this population whose speed exceeded 105 wpm? d. Suppose that two typists are independently selected. What is the probability that both their speeds exceed 75 wpm? e. Suppose that special training is to be made available to the slowest \(20 \%\) of the typists. What typing speeds would qualify individuals for this training? (Hint: See Example 6.23\()\)

The probability distribution of \(x,\) the number of defective tires on a randomly selected automobile checked at a certain inspection station, is given in the following table: $$ \begin{array}{lccccc} x & 0 & 1 & 2 & 3 & 4 \\ p(x) & 0.54 & 0.16 & 0.06 & 0.04 & 0.20 \end{array} $$ The mean value of \(x\) is \(\mu_{x}=1.2\). Calculate the values of \(\sigma_{x}^{2}\) and \(\sigma_{x}\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.