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Chapter 6: Problem 22

Television viewing reached a new high when the Nielsen Company reported a mean daily viewing time of 8.35 hours per household (USA Today, November 11,2009 ). Use a normal probability distribution with a standard deviation of 2.5 hours to answer the following questions about daily television viewing per household. a. What is the probability that a household views television between 5 and 10 hours a day? b. How many hours of television viewing must a household have in order to be in the top \(3 \%\) of all television viewing households? c. What is the probability that a household views television more than 3 hours a day?

Short Answer

Expert verified
a) 0.6553, b) 12.57 hours, c) 0.9838.

Step by step solution

01

Identify the Parameters

We are given a normal distribution with a mean \( \mu = 8.35 \) hours and a standard deviation \( \sigma = 2.5 \) hours.
02

Part (a): Define the Range and Standardize

For a household that views television between 5 and 10 hours a day, we need to find \( P(5 < X < 10) \). To calculate this, we first standardize using the formula \( Z = \frac{X - \mu}{\sigma} \): - For \( X = 5 \), \( Z_1 = \frac{5 - 8.35}{2.5} = -1.34 \).- For \( X = 10 \), \( Z_2 = \frac{10 - 8.35}{2.5} = 0.66 \).
03

Part (a): Compute the Probability

Using the standard normal distribution table, or a calculator, find the probabilities: - \( P(Z < -1.34) \approx 0.0901 \).- \( P(Z < 0.66) \approx 0.7454 \).Thus, \( P(5 < X < 10) = P(Z < 0.66) - P(Z < -1.34) \approx 0.7454 - 0.0901 = 0.6553 \).
04

Part (b): Find Z-score for Top 3%

For a household to be in the top 3%, we need the Z-score that corresponds to the 97th percentile (since 100% - 3% = 97%). From the Z-table or calculator, \( Z \approx 1.88 \).
05

Part (b): Convert Z-score to Time

Convert the Z-score into hours with the formula: \[ X = \mu + Z \times \sigma = 8.35 + 1.88 \times 2.5 = 12.57 \] A household must view television for approximately 12.57 hours to be in the top 3%.
06

Part (c): Calculate Z for More than 3 Hours

For a household viewing more than 3 hours, find \( P(X > 3) \). First, calculate the Z-score:\[ Z = \frac{3 - 8.35}{2.5} = -2.14 \]
07

Part (c): Compute the Probability

Using the Z-table or calculator, \( P(Z < -2.14) \approx 0.0162 \). Since we need \( P(X > 3) \), compute:\[ P(X > 3) = 1 - P(Z < -2.14) \approx 1 - 0.0162 = 0.9838 \].

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Key Concepts

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

Probability Calculation
Probability calculation is a fundamental concept used to determine how likely events are to occur within a given dataset. In this exercise, the focus is on finding the probability of certain intervals and conditions using the normal distribution as a model for daily TV viewing times.
You use the normal distribution because many variables in nature follow this pattern, where most observations cluster around the mean and the probability decreases symmetrically as you move away from the mean.
  • To calculate probabilities using a normal distribution, we typically rely on Z-scores, which map raw scores onto a standard normal distribution.
  • For example, finding the probability of watching TV between 5 and 10 hours per day involves standardizing these values to Z-scores.
  • You then use a Z-table, or similar tools like statistical software, to find the probability between these scores, reflecting the likelihood of the event.

Overall, probability calculation helps in making informed decisions based on statistical models, expanding our understanding beyond mere averages.
Z-Score
The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. Essentially, it indicates how many standard deviations an element is from the mean.
By converting a value to a Z-score, you can see its position within the normal distribution, making it easier to compute probabilities.
  • The formula for Z-score is given by \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
  • In this exercise, for a household viewing TV between 5 and 10 hours, these respective raw scores are converted to Z-scores of -1.34 and 0.66.
  • The conversion allows you to use the standard normal distribution to find probabilities between these Z-values.

Understanding Z-scores is crucial as they enable you to work with different datasets under a standardized framework, providing a clearer insight into the data's characteristics.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In a normal distribution, it defines how spread out the data points are around the mean.
A low standard deviation means that data points are generally close to the mean, whereas a high standard deviation indicates that the data points are spread out over a wider range.
  • In the context of this exercise, a standard deviation of 2.5 hours shows the average spread of household TV viewing times from the mean of 8.35 hours.
  • This measure is crucial because it affects how probabilities are calculated for given intervals of TV viewing durations.
  • The standard deviation directly influences the positioning of Z-scores, impacting the probability calculations derived from these scores.

Understanding standard deviation helps in grasping how much the data varies and is essential for interpreting results within the context of normal distribution.

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Most popular questions from this chapter

The average stock price for companies making up the S\&P 500 is \(\$ 30,\) and the standard deviation is \(\$ 8.20\) (Business Week, Special Annual Issue, Spring 2003 ). Assume the stock prices are normally distributed. a. What is the probability that a company will have a stock price of at least \(\$ 40 ?\) b. What is the probability that a company will have a stock price no higher than \(\$ 20 ?\) c. How high does a stock price have to be to put a company in the top \(10 \% ?\)

The time between arrivals of vehicles at a particular intersection follows an exponential probability distribution with a mean of 12 seconds. a. Sketch this exponential probability distribution. b. What is the probability that the arrival time between vehicles is 12 seconds or less? c. What is the probability that the arrival time between vehicles is 6 seconds or less? d. What is the probability of 30 or more seconds between vehicle arrivals?

Delta Air Lines quotes a flight time of 2 hours, 5 minutes for its flights from Cincinnati to Tampa. Suppose we believe that actual flight times are uniformly distributed between 2 hours and 2 hours, 20 minutes. a. Show the graph of the probability density function for flight time. b. What is the probability that the flight will be no more than 5 minutes late? c. What is the probability that the flight will be more than 10 minutes late? d. What is the expected flight time?

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

The website for the Bed and Breakfast Inns of North America gets approximately seven visitors per minute. Suppose the number of website visitors per minute follows a Poisson probability distribution. a. What is the mean time between visits to the website? b. Show the exponential probability density function for the time between website visits. c. What is the probability that no one will access the website in a 1-minute period? d. What is the probability that no one will access the website in a 12 -second period?
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