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According to media research, the typical American listened to 195 hours of music in the last year. This is down from 290 hours 4 years earlier. Dick Trythal is a big country and western music fan. He listens to music while working around the house, reading, and riding in his truck. Assume the number of hours spent listening to music follows a normal probability distribution with a standard deviation of 8.5 hours. a. If Dick is in the top \(1 \%\) in terms of listening time, how many hours did he listen last year? b. Assume that the distribution of times 4 years earlier also follows the normal probability distribution with a standard deviation of 8.5 hours. How many hours did the \(1 \%\) who listen to the least music actually listen?

Short Answer

Expert verified
a. Approximately 214.8 hours; b. Approximately 270.2 hours.

Step by step solution

01

Identification

Identify what is given in the question. For Dick's listening time, the problem states that he is in the top 1%. The mean (average) listening time is now 195 hours with a standard deviation of 8.5 hours. We want to find the listening time corresponding to the top 1%. For the listeners from 4 years ago, we take the mean as 290 hours with the same standard deviation. We want to find the time corresponding to the bottom 1% in this distribution.
02

Convert Top 1% to a Z-score

The top 1% corresponds to a cumulative probability of 0.99 in a normal distribution. Using standard normal distribution tables or a calculator, find the Z-score for a cumulative probability of 0.99, which is approximately 2.33.
03

Calculate Dick's Listening Time

Using the Z-score equation, we know \[ x = ext{mean} + Z imes ext{standard deviation} \]Substituting the known values, \[ x = 195 + 2.33 imes 8.5 \]Calculate the value to find Dick's actual listening time.
04

Convert Bottom 1% to a Z-score

For the bottom 1%, we need the cumulative probability of 0.01. The Z-score for this probability is approximately -2.33.
05

Calculate 4 Years Ago Listening Time

Using the Z-score equation, \[ x = ext{mean} + Z imes ext{standard deviation} \]Substituting the known values for 4 years ago,\[ x = 290 + (-2.33) imes 8.5 \]Calculate this to find the number of hours for the bottom 1%.

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Key Concepts

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

Normal Distribution
In statistics, the normal distribution is a crucial concept often referred to as the bell curve due to its bell-shaped appearance when graphed. It represents a probability distribution where most occurrences take place near the mean, and the probabilities taper off symmetrically on both sides. This makes it useful in various fields, such as finance, science, and social sciences, to model data. The defining characteristics of a normal distribution include:
  • The mean, median, and mode are all equal.
  • It is perfectly symmetric around its mean.
  • The bell shape is characterized by how wide or narrow it is, which is determined by the standard deviation.
In problems involving normal distribution, like the one we have, we often wish to find the proportion of data that falls within a certain range. This is achieved by converting the data points into a standardized form using Z-scores.
Z-score
A Z-score, or standard score, is a statistical measure that describes a data point's position relative to the mean of the data set. By transforming data into Z-scores, we can determine how many standard deviations away a particular point is from the mean. This is extremely useful for comparing different data points within the same distribution or across different distributions.Here's how you calculate a Z-score:
  • Identify the mean (\( \ar{x} \)) of the data set.
  • Determine the individual data point (\( x \)) for which the Z-score is needed.
  • Use the formula: \( Z = \frac{x - \text{mean}}{\text{standard deviation}} \)
In the exercise provided, finding the Z-score corresponding to the top and bottom 1% of listeners involved identifying these cut-off points. A table or calculator is used to find the required Z-scores, which are then used in calculations to determine the exact listening hours.
Probability Distribution
A probability distribution describes how probabilities are distributed over the values of a random variable. This is a fundamental concept in statistics as it provides a mathematical function that gives the probabilities of occurrence of different possible outcomes for an experiment. For a normal probability distribution, the probabilities are distributed in a symmetrical pattern, clustering around the mean. The properties include:
  • The total area under the curve equals 1, representing 100% probability.
  • Probabilities for intervals can be calculated by determining the area under the curve for that interval.
In real-world scenarios, probability distributions can help to predict outcomes and make informed decisions. For instance, in the context of our exercise, knowing the probability distribution of music listening hours allows us to determine where individuals like Dick rank in terms of small or large amounts of listening time.
Standard Deviation
Standard deviation is a statistic that measures the dispersion or spread in a set of data. It's a crucial element in determining how much variability there is between observations in a dataset.Key points about standard deviation:
  • A low standard deviation indicates that the data points tend to be close to the mean.
  • A high standard deviation means the data points are spread out over a large range of values.
Calculating standard deviation for a data set involves computing the square root of the average squared deviation from the mean:\[ \text{Standard Deviation} = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(x_i - \bar{x})^2} \]where \(x_i\) represents each value in the data set, \(\bar{x}\) is the mean of the data set, and \(N\) is the number of observations.In the given problem, the standard deviation helps assess listener behavior over a year by indicating the spread of listening hours in relation to the mean.

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Most popular questions from this chapter

Best Electronics Inc. offers a "no hassle" returns policy. The daily number of customers returning items follows the normal distribution. The mean number of customers returning items is 10.3 per day and the standard deviation is 2.25 per day. a. For any day, what is the probability that eight or fewer customers returned items? b. For any day, what is the probability that the number of customers returning items is between 12 and \(14 ?\) c. Is there any chance of a day with no customer returns?

List the major characteristics of a normal probability distribution.

A normal population has a mean of 12.2 and a standard deviation of \(2.5 .\) a. Compute the \(z\) value associated with 14.3 . b. What proportion of the population is between 12.2 and \(14.3 ?\) c. What proportion of the population is less than \(10.0 ?\)

The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is \(12,200 .\) The distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation is 820 pages. The manufacturer wants to provide guidelines to potential customers as to how long they can expect a cartridge to last. How many pages should the manufacturer advertise for each cartridge if it wants to be correct \(99 \%\) of the time?

Most four-year automobile leases allow up to 60,000 miles. If the lessee goes beyond this amount, a penalty of 20 cents per mile is added to the lease cost. Suppose the distribution of miles driven on four-year leases follows the normal distribution. The mean is 52,000 miles and the standard deviation is 5,000 miles. a. What percent of the leases will yield a penalty because of excess mileage? b. If the automobile company wanted to change the terms of the lease so that \(25 \%\) of the leases went over the limit, where should the new upper limit be set? c. One definition of a low-mileage car is one that is 4 years old and has been driven less than 45,000 miles. What percent of the cars returned are considered low-mileage?

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