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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. \(P(x \leq 2) = P(x=0) + P(x=1) + P(x=2) = 0.10 + 0.15 + 0.20 = 0.45\).

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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.

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Most popular questions from this chapter

Example 6.14 gave the probability distributions shown below for \(x=\) number of flaws in a randomly selected glass panel from Supplier 1 \(y=\) number of flaws in a randomly selected glass panel from Supplier 2 for two suppliers of glass used in the manufacture of flat screen TVs. If the manufacturer wanted to select a single supplier for glass panels, which of these two suppliers would you recommend? Justify your choice based on consideration of both center and variability. $$ \begin{array}{lcccclcccc} \boldsymbol{x} & 0 & 1 & 2 & 3 & \boldsymbol{y} & 0 & 1 & 2 & 3 \\ \boldsymbol{p}(\boldsymbol{x}) & 0.4 & 0.3 & 0.2 & 0.1 & \boldsymbol{p}(\boldsymbol{y}) & 0.2 & 0.6 & 0.2 & 0 \end{array} $$

An appliance dealer sells three different models of freezers having \(13.5,15.9,\) and 19.1 cubic feet of storage space. Let \(x=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(x\) has the following probability distribution: $$ \begin{array}{lrrr} x & 13.5 & 15.9 & 19.1 \\ p(x) & 0.2 & 0.5 & 0.3 \end{array} $$ a. Calculate the mean and standard deviation of \(x\). (Hint: See Example 6.15\()\) b. Give an interpretation of the mean and standard deviation of \(x\) in the context of observing the outcomes of many purchases.

Consider the variable \(x=\) time required for a college student to complete a standardized exam. Suppose that for the population of students at a particular university, the distribution of \(x\) is well approximated by a normal curve with mean 45 minutes and standard deviation 5 minutes. a. If 50 minutes is allowed for the exam, what proportion of students at this university would be unable to finish in the allotted time? b. How much time should be allowed for the exam if you wanted \(90 \%\) of the students taking the test to be able to finish in the allotted time? c. How much time is required for the fastest \(25 \%\) of all tudents to complete the exam

A gasoline tank for a certain car is designed to hold 15 gallons of gas. Suppose that the random variable \(x=\) actual capacity of a randomly selected tank has a distribution that is well approximated by a normal curve with mean 15.0 gallons and standard deviation 0.1 gallon. a. What is the probability that a randomly selected tank will hold at most 14.8 gallons? b. What is the probability that a randomly selected tank will hold between 14.7 and 15.1 gallons? c. If two such tanks are independently selected, what is the probability that both hold at most 15 gallons?

Suppose that fuel efficiency (miles per gallon, mpg) for a particular car model under specified conditions is normally distributed with a mean value of \(30.0 \mathrm{mpg}\) and a standard deviation of \(1.2 \mathrm{mpg}\). a. What is the probability that the fuel efficiency for a randomly selected car of this model is between 29 and \(31 \mathrm{mpg} ?\) b. Would it surprise you to find that the efficiency of a randomly selected car of this model is less than \(25 \mathrm{mpg} ?\) c. If three cars of this model are randomly selected, what is the probability that each of the three have efficiencies exceeding \(32 \mathrm{mpg}\) ? d. Find a number \(x^{*}\) such that \(95 \%\) of all cars of this model have efficiencies exceeding \(x^{*}\left(\right.\) i.e., \(\left.P\left(x>x^{*}\right)=0.95\right)\).

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