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Chapter 6: Problem 35

A local television station sells 15 -second, 30 -second, and 60 -second advertising spots. Let \(x\) denote the length of a randomly selected commercial appearing on this station, and suppose that the probability distribution of \(x\) is given by the following table: $$ \begin{array}{lrrr} x & 15 & 30 & 60 \\ p(x) & 0.1 & 0.3 & 0.6 \end{array} $$ What is the mean length for commercials appearing on this station?

Short Answer

Expert verified
The mean length for commercials appearing on the station is 46.5 seconds.

Step by step solution

01

Identify the Variables and their Probabilities

From the given problem, we have three possible lengths for the commercial, 15, 30, and 60 seconds and the associated probabilities are 0.1, 0.3, and 0.6 respectively. These can be represented as pairs: (15, 0.1), (30, 0.3), (60, 0.6).
02

Apply the Expected Value Formula

We calculate the mean (expected value) using the formula \(\mu = \Sigma [x * P(x)]\). Plug the values from Step 1 into the formula: \(\mu = (15 * 0.1) + (30 * 0.3) + (60 * 0.6)\)
03

Calculate the Mean

Carry out the multiplication and addition operations to find the mean: \( \mu = 1.5 + 9 + 36 = 46.5\)

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Key Concepts

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

Probability Distribution
Understanding the probability distribution is critical to analyzing data and making predictions about future events. It is a statistical function that describes all the possible values and likelihoods that a random variable can take within a given range. A probability distribution is like a map that guides us in the landscape of probability, showing where we're likely to find the values of a random variable and with what frequency.

In our exercise, the random variable represents the length of a commercial, and the probability distribution is given by the table, linking each commercial length to its probability. For example, we see that a 60-second commercial is most likely to occur, as it has the highest probability (0.6). When we sum all the probabilities, they should equal 1, confirming that our distribution covers all possible outcomes of the random variable.
Mean
The mean, often known as the expected value in the context of probability, is the long-run average value of repetitions of the experiment it represents. Simply put, it's what you would anticipate as the outcome over an extended period or over many trials. It takes into account all possible values and their probabilities, making it a fundamental measure of the central tendency of a probability distribution.

In calculating the mean length of commercials in our example, we multiplied each length of the commercial by its respective probability and added the values together. This calculation provided us with the average length of a commercial airing on the television station, which is 46.5 seconds.
Random Variable
A random variable is a variable whose possible values are numerical outcomes of a random phenomenon. There are two types of random variables: discrete and continuous. Discrete random variables, like the one in our textbook example, have a countable number of possibilities (e.g., the lengths of commercials). Continuous random variables, on the other hand, can take an infinite number of values within a range.

In the context of our exercise, the random variable 'x' represents the possible lengths of commercials (15, 30, and 60 seconds), which can be observed with certain probabilities. Random variables are the foundations on which probability distributions are built, and understanding them is essential for working with any statistical model.

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Most popular questions from this chapter

A coin is flipped 25 times. Let \(x\) be the number of flips that result in heads (H). Consider the following rule for deciding whether or not the coin is fair: Judge the coin fair if \(8 \leq x \leq 17\). Judge the coin biased if either \(x \leq 7\) or \(x \geq 18\). a. What is the probability of judging the coin biased when it is actually fair? b. Suppose that a coin is not fair and that \(P(\mathrm{H})=0.9\). What is the probability that this coin would be judged fair? What is the probability of judging a coin fair if \(P(\mathrm{H})=0.1 ?\) c. What is the probability of judging a coin fair if \(P(\mathrm{H})=0.6 ?\) if \(P(\mathrm{H})=0.4 ?\) Why are these probabilities large compared to the probabilities in Part (b)? d. What happens to the "error probabilities" of Parts (a) and \((b)\) if the decision rule is changed so that the coin is judged fair if \(7 \leq x \leq 18\) and unfair otherwise? Is this a better rule than the one first proposed? Explain.

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?

The article "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants" (Water Research [1984]: \(1169-1174\) ) suggests the uniform distribution on the interval from 7.5 to 20 as a model for \(x=\) depth (in centimeters) of the bioturbation layer in sediment for a certain region. a. Draw the density curve for \(x\). b. What is the height of the density curve? c. What is the probability that \(x\) is at most \(12 ?\) d. What is the probability that \(x\) is between 10 and 15 ? Between 12 and 17 ? Why are these two probabilities equal?

Suppose \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. The probability distribution of \(y\) is as follows: $$ \begin{array}{lccccl} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & ? \end{array} $$ a. Only \(y\) values of \(0,1,2,3,\) and 4 have probabilities greater than 0 . What is \(p(4)\) ? b. How would you interpret \(p(1)=0.20 ?\) c. Calculate \(P(y \leq 2)\), the probability that the carton contains at most two broken eggs, and interpret this probability. d. Calculate \(P(y<2),\) the probability that the carton contains fewer than two broken eggs. Why is this smaller than the probability in Part (c)? e. What is the probability that the carton contains exactly 10 unbroken eggs? f. What is the probability that at least 10 eggs are unbroken?

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 \(z > 2.50)\)
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