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A mail-order computer software business has six telephone lines. Let \(x\) denote the number of lines in use at a specified time. The probability distribution of \(x\) is as follows: $$ \begin{array}{lrrrrrrr} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ p(x) & .10 & .15 & .20 & .25 & .20 & .06 & .04 \end{array} $$ 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 for the events are: a. 0.70, b. 0.45, c. 0.55, d. 0.71, e. 0.45, f. 0.45.

Step by step solution

01

Creating a Probability Table

Initially, make a note that \(x\) can have values from 0 to 6 and the corresponding probabilities are 0.10, 0.15, 0.20, 0.25, 0.20, 0.06, 0.04, respectively. The sum of these probabilities should total to 1.
02

Calculating the probability for event a

Event a is defined as 'At most three lines are in use'. This means \(x\) can be 0, 1, 2, or 3. We calculate the probability by summing up the probabilities for each possible value of \(x\): \(p(x=0) + p(x=1) + p(x=2) + p(x=3) = 0.10 + 0.15 + 0.20 + 0.25 = 0.70\).
03

Calculating the probability for event b

Event b is defined as 'Fewer than three lines are in use'. This means \(x\) can be 0, 1, or 2. Therefore, we add the probabilities for these values: \(p(x=0) + p(x=1) + p(x=2) = 0.10 + 0.15 + 0.20 = 0.45\).
04

Calculating the probability for event c

Event c is 'At least three lines are in use'. This means \(x\) can be 3, 4, 5, or 6. So, the probability is \(p(x=3) + p(x=4) + p(x=5) + p(x=6) = 0.25 + 0.20 + 0.06 + 0.04 = 0.55\).
05

Calculating the probability for event d

Event d is 'Between two and five lines (inclusive) are in use'. This means \(x\) can be 2, 3, 4 or 5. Therefore the probability is \(p(x=2) + p(x=3) + p(x=4) + p(x=5) = 0.20 + 0.25 + 0.20 + 0.06 = 0.71\).
06

Calculating the probability for event e

Event e is 'Between two and four lines (inclusive) are not in use'. This means \(x\) can be 0, 1 or 2. So, the probability is \(p(x=0) + p(x=1) + p(x=2) = 0.10 + 0.15 + 0.20 = 0.45\).
07

Calculating the probability for event f

Event f is 'At least four lines are not in use'. This means \(x\) can be 0, 1, or 2. So, probability is \(p(x=0) + p(x=1) + p(x=2) = 0.10 + 0.15 + 0.20 = 0.45\).

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Key Concepts

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

Discrete Random Variables
Taking the first dive into probability, let's talk about discrete random variables. These are a cornerstone concept and relate to countable outcomes. Consider a call center with multiple phone lines; in this case, the number of lines in use at any given time is a prime example of a discrete random variable. It can take on finite or countable values, such as 0, 1, 2, all the way up to 6.

In essence, a discrete random variable, which we can denote as 'X', represents distinct outcomes of a random experiment, like flipping a coin, where 'X' can be 0 for tails and 1 for heads. Similarly, for our mail-order computer software business with six telephone lines, 'X' could be 0 when no lines are in use or 6 when all lines are engaged, with each state having a certain probability linked to it. Each possible state of 'X' is a fundamental piece of the puzzle in understanding the behaviour of systems influenced by chance.
Probability Mass Function
So, how do we quantify the likelihood of each outcome for our discrete random variable? Enter the probability mass function (PMF), symbolized as p(x). It is the function that gives us the probability that a discrete random variable is exactly equal to some value. For example, in our textbook problem, the PMF tells us the probability of 0 lines being in use is 0.10, 1 line is 0.15, and so on.

The PMF is interesting because it helps us create a complete picture of all possible outcomes and their probabilities. Mathematically, for a PMF to be valid, it must satisfy two conditions: the sum of all probabilities must equal 1, and the probability for each individual outcome must be between 0 and 1. This makes perfect sense because the probabilities of all possible outcomes must cover every scenario (hence totaling to 1), and no single outcome can be a certainty or an impossibility (hence the 0 to 1 range).
Cumulative Probability
Now that we have our PMF defined, we can explore cumulative probability. This entails looking at the probability of the event where our discrete random variable 'X' is less than or equal to a certain value. It's like summarizing the probabilities up to a point. This cumulative aspect is captured by the cumulative distribution function (CDF), which is obtained by summing up the probabilities from the start of the distribution up to a certain point 'x'.

In our exercise, when asked to find the probability that at most three lines are in use, we found it by adding up the probabilities up to and including when three lines are used—essentially, the CDF for 3. It is a powerful tool that helps students handle questions such as 'What's the probability of getting three or fewer calls?' without having to consider each possible outcome separately. The cumulative probability can answer 'at most', 'less than', 'greater than', or 'at least' types of questions, all vital in understanding the probability distribution of a discrete random variable.

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Most popular questions from this chapter

In a study of warp breakage during the weaving of fabric (Technometrics [1982]: 63), 100 pieces of yarn were tested. The number of cycles of strain to breakage was recorded for each yarn sample. The resulting data are given in the following table: $$ \begin{array}{rrrrrrrrrr} 86 & 146 & 251 & 653 & 98 & 249 & 400 & 292 & 131 & 176 \\ 76 & 264 & 15 & 364 & 195 & 262 & 88 & 264 & 42 & 321 \\ 180 & 198 & 38 & 20 & 61 & 121 & 282 & 180 & 325 & 250 \\ 196 & 90 & 229 & 166 & 38 & 337 & 341 & 40 & 40 & 135 \\ 597 & 246 & 211 & 180 & 93 & 571 & 124 & 279 & 81 & 186 \\ 497 & 182 & 423 & 185 & 338 & 290 & 398 & 71 & 246 & 185 \\ 188 & 568 & 55 & 244 & 20 & 284 & 93 & 396 & 203 & 829 \\ 239 & 236 & 277 & 143 & 198 & 264 & 105 & 203 & 124 & 137 \\ 135 & 169 & 157 & 224 & 65 & 315 & 229 & 55 & 286 & 350 \\ 193 & 175 & 220 & 149 & 151 & 353 & 400 & 61 & 194 & 188 \end{array} $$ a. Construct a frequency distribution using the class intervals 0 to \(<100,100\) to \(<200\), and so on. b. Draw the histogram corresponding to the frequency distribution in Part (a). How would you describe the shape of this histogram? c. Find a transformation for these data that results in a more symmetric histogram than what you obtained in Part (b).

A sporting goods store has a special sale on three brands of tennis balls - call them D, P, and W. Because the sale price is so low, only one can of balls will be sold to each customer. If \(40 \%\) of all customers buy Brand \(\mathrm{W}\), \(35 \%\) buy Brand \(\mathrm{P}\), and \(25 \%\) buy Brand \(\mathrm{D}\) and if \(x\) is the number among three randomly selected customers who buy Brand \(\mathrm{W}\), what is the probability distribution of \(x\) ?

Thirty percent of all automobiles undergoing an emissions inspection at a certain inspection station fail the inspection. a. Among 15 randomly selected cars, what is the probability that at most 5 fail the inspection? b. Among 15 randomly selected cars, what is the probability that between 5 and 10 (inclusive) fail to pass inspection? c. Among 25 randomly selected cars, what is the mean value of the number that pass inspection, and what is the standard deviation of the number that pass inspection? d. What is the probability that among 25 randomly selected cars, the number that pass is within 1 standard deviation of the mean value?

Accurate labeling of packaged meat is difficult because of weight decrease resulting from moisture loss (defined as a percentage of the package's original net weight). Suppose that moisture loss for a package of chicken breasts is normally distributed with mean value \(4.0 \%\) and standard deviation \(1.0 \% .\) (This model is suggested in the paper "Drained Weight Labeling for Meat and Poultry: An Economic Analysis of a Regulatory Proposal," Journal of Consumer Affairs [1980]: 307-325.) Let \(x\) denote the moisture loss for a randomly selected package. a. What is the probability that \(x\) is between \(3.0 \%\) and \(5.0 \%\) ? b. What is the probability that \(x\) is at most \(4.0 \%\) ? c. What is the probability that \(x\) is at least \(7.0 \%\) ? d. Find a number \(z^{*}\) such that \(90 \%\) of all packages have moisture losses below \(z^{*} \%\). e. What is the probability that moisture loss differs from the mean value by at least \(1 \%\) ?

To assemble a piece of furniture, a wood peg must be inserted into a predrilled hole. Suppose that the diameter of a randomly selected peg is a random variable with mean \(0.25\) in. and standard deviation \(0.006\) in. and that the diameter of a randomly selected hole is a random variable with mean \(0.253\) in. and standard deviation \(0.002\) in. Let \(x_{1}=\) peg diameter, and let \(x_{2}=\) denote hole diameter. a. Why would the random variable \(y\), defined as \(y=\) \(x_{2}-x_{1}\), be of interest to the furniture manufacturer? b. What is the mean value of the random variable \(y\) ? c. Assuming that \(x_{1}\) and \(x_{2}\) are independent, what is the standard deviation of \(y\) ? d. Is it reasonable to think that \(x_{1}\) and \(x_{2}\) are independent? Explain. e. Based on your answers to Parts (b) and (c), do you think that finding a peg that is too big to fit in the predrilled hole would be a relatively common or a relatively rare occurrence? Explain.

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