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Chapter 6: Problem 6

On average, 30 -minute television sitcoms have 22 minutes of programming (CNBC, February 23,2006 . Assume that the probability distribution for minutes of programming can be approximated by a uniform distribution from 18 minutes to 26 minutes. a. What is the probability that a sitcom will have 25 or more minutes of programming? b. What is the probability that a sitcom will have between 21 and 25 minutes of programming? c. What is the probability that a sitcom will have more than 10 minutes of commercials or other nonprogramming interruptions?

Short Answer

Expert verified
a. \(\frac{1}{8}\), b. \(\frac{1}{2}\), c. \(\frac{1}{4}\).

Step by step solution

01

Understanding Uniform Distribution

A uniform distribution is a probability distribution where all outcomes are equally likely. For this exercise, we have a uniform distribution of programming time from 18 to 26 minutes. This means every minute within this range is equally likely for a sitcom's programming time.
02

Calculating Total Range of Distribution

The total range of the uniform distribution is the difference between the maximum and minimum values. Thus, it is calculated as:\[ \text{Range} = 26 - 18 = 8 \text{ minutes}\]This is the total range of possible programming minutes.
03

Calculating Probability for Part (a)

For part (a), we want to determine the probability that programming time is 25 minutes or more. The range for this event is from 25 to 26 minutes. Therefore, the probability can be calculated as:\[P(X \geq 25) = \frac{26 - 25}{26 - 18} = \frac{1}{8}\]This gives the probability of programming being 25 minutes or more.
04

Calculating Probability for Part (b)

For part (b), we need the probability that programming time is between 21 and 25 minutes. Thus we calculate it as:\[P(21 \leq X \leq 25) = \frac{25 - 21}{26 - 18} = \frac{4}{8} = \frac{1}{2}\]This is the probability of programming being between 21 and 25 minutes.
05

Calculating Probability for Part (c)

For part (c), we first determine the maximum time for nonprogramming interruptions, which would be 30 minutes minus the minimum programming time (18 minutes), resulting in 12 minutes of interruptions. We then find the probability that programming is less than 20 minutes so that the interruptions are more than 10 minutes.\[P(X < 20) = \frac{20 - 18}{26 - 18} = \frac{2}{8} = \frac{1}{4}\]This is the probability of having more than 10 minutes of interruptions.

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Key Concepts

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

Probability Calculation
Probability calculation involves determining the likelihood of a specific event happening within a defined set of possibilities. In a uniform distribution, like the one used in our exercise, all outcomes within a certain range are equally probable. This means that if we want to calculate the probability of an event involving time, such as a sitcom having at least a certain number of minutes of programming, we only need to focus on figuring out the ratio of the favorable time interval to the total possible time interval.
To calculate this probability, we first identify the specific range of minutes that interest us. For instance, in part (a) of our exercise, we're interested in programming that lasts 25 minutes or more. We then calculate how many minutes fall within this range and divide by the total range of all possible outcomes in the uniform distribution. This method leverages the unique properties of a uniform distribution, where the probability over any interval is the interval's length divided by the entire distribution length.
Probability Distribution
A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. The uniform distribution, a simple type of distribution, posits that all events are equally probable within a defined range. This results in a flat, evenly distributed curve, representing equal likelihoods.
For our exercise, we are dealing with a uniform distribution of programming time for sitcoms ranging from 18 to 26 minutes. This means every minute within this interval has the same chance of occurring, highlighting that not one particular programming length is favored over another.
Using this principle, we can calculate probabilities by examining the proportion of the interval of interest relative to the entire distribution range. For example, when checking the probability of programming time being between 21 and 25 minutes, we see that this represents a segment of the overall range. The calculation involves the simple mathematics of dividing the length of the interest range by the total range length, demonstrating the simplicity and elegance of uniform probability distributions in statistics.
Statistics
Statistics is the science that deals with the collection, analysis, interpretation, and presentation of data. Uniform distribution is just one of many tools in statistics that help us understand data patterns. By observing the uniform distribution, statistics allows us to predict outcomes and make informed decisions based on probability.
In our exercise, statistics helps us analyze the likely duration of programming and related interruptions through probability calculations. These calculations help us infer part of the experience of watching TV sitcoms by giving insights into programming trends and patterns.
Statistics, using uniform distribution, simplifies complex real-world phenomenons into predictable models, offering a sense of certainty within variability. It broadens our toolkit, providing structured ways to quantify probability, which is essential for various practical applications, from business forecasts to scientific research, thereby reinforcing its pivotal role in daily decision-making processes.

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Most popular questions from this chapter

Is lack of sleep causing traffic fatalities? A study conducted under the auspices of the National Highway Traffic Safety Administration found that the average number of fatal crashes caused by drowsy drivers each year was 1550 ( Business Week , January 26,2004 ). Assume the annual number of fatal crashes per year is normally distributed with a standard deviation of 300 . a. What is the probability of fewer than 1000 fatal crashes in a year? b. What is the probability that the number of fatal crashes will be between 1000 and 2000 for a year? c. For a year to be in the upper \(5 \%\) with respect to the number of fatal crashes, how many fatal crashes would have to occur?

Assume a binomial probability distribution has \(p=.60\) and \(n=200\) a. What are the mean and standard deviation? b. Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain. c. What is the probability of 100 to 110 successes? d. What is the probability of 130 or more successes? e. What is the advantage of using the normal probability distribution to approximate the binomial probabilities? Use part (d) to explain the advantage.

Given that \(z\) is a standard normal random variable, compute the following probabilities. a. \(\quad P(0 \leq z \leq .83)\) b. \(\quad P(-1.57 \leq z \leq 0)\) c. \(\quad P(z>.44)\) d. \(\quad P(z \geq-.23)\) e. \(\quad P(z<1.20)\) f. \(\quad P(z \leq-.71)\)

The Information Systems Audit and Control Association surveyed office workers to learn about the anticipated usage of office computers for personal holiday shopping (USA Today, November 11,2009 ). Assume that the number of hours a worker spends doing holiday shopping on an office computer follows an exponential distribution. a. The study reported that there is a .53 probability that a worker uses an office computer for holiday shopping 5 hours or less. Is the mean time spent using an office computer for holiday shopping closest to \(5.8,6.2,6.6,\) or 7 hours? b. Using the mean time from part (a), what is the probability that a worker uses an office computer for holiday shopping more than 10 hours? c. What is the probability that a worker uses an office computer for holiday shopping between 4 and 8 hours?

The random variable \(x\) is known to be uniformly distributed between 1.0 and 1.5 a. Show the graph of the probability density function. b. Compute \(P(x=1.25)\) c. Compute \(P(1.0 \leq x \leq 1.25)\) d. Compute \(P(1.20
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