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An energy study in Gainesville, Florida, found that in March 2006 , household use of electricity had a mean of 673 and a standard deviation of 556 kilowatthours. (Source: Data from Todd Kamhoot, Gainesville Regional Utilities.) a. Suppose the distribution of energy use was normal. Using a table, calculator, or software that can give normal probabilities, find the proportion of households with electricity use greater than 1000 kilowatt- hours. b. Based on the mean and standard deviation given, do you think that the distribution of energy use actually is normal? Why or why not?

Short Answer

Expert verified
a. About 27.76% of households use more than 1000 kWh. b. The large standard deviation relative to the mean suggests the distribution may not be normal.

Step by step solution

01

Understand the Problem

We need to determine the proportion of households whose electricity usage is greater than 1000 kWh, assuming the usage follows a normal distribution with mean (μ) = 673 kWh and standard deviation (σ) = 556 kWh. We will then evaluate the appropriateness of assuming this distribution is normal.
02

Standardize the Value

To find the proportion of households using more than 1000 kWh, we first convert 1000 kWh into a z-score using the formula: \( z = \frac{X - \mu}{\sigma} \), where \( X = 1000, \mu = 673, \sigma = 556 \). Thus, \( z = \frac{1000 - 673}{556} = 0.586 \).
03

Use Z-table or Software to Find Probability

Using the z-score obtained, we look up this value in a standard normal distribution table or use software to determine the probability. The z-score of 0.586 corresponds to a cumulative probability of approximately 0.7224. This means 72.24% of households use less than 1000 kWh.
04

Calculate Proportion Greater Than 1000 kWh

To find the proportion of households using more than 1000 kWh, we subtract the cumulative probability from 1: \( 1 - 0.7224 = 0.2776 \). Thus, approximately 27.76% of households use more than 1000 kWh.
05

Evaluate the Normality Assumption

Considering the large standard deviation relative to the mean (556 compared to 673), this suggests a highly variable data set. Such a large standard deviation implies that the distribution may not be perfectly normal, as a normal distribution typically has a smaller relative dispersion. Also, energy use might be influenced by external factors that can cause deviations from normality.

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Key Concepts

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

Standard Deviation
The standard deviation is a key concept in statistics that measures the amount of variation or dispersion within a set of data. In simple terms, it tells us how spread out the values in a data set are around the mean (average).
For example, if the standard deviation is small, the data points tend to be close to the mean value, indicating consistency in the data. On the other hand, a large standard deviation means that the data points are spread out over a wide range of values, indicating more variability.
  • In the context of the energy use study in Gainesville, the standard deviation of 556 kWh compares quite closely to the mean of 673 kWh, showing substantial variation in household electricity usage.
  • This large standard deviation suggests that while some households are near the average usage, others use significantly more or less electricity.
  • Understanding standard deviation helps in assessing whether the data is clustered or dispersed, which is vital for evaluating the normality assumption of any dataset.
Knowing the variability helps in making informed decisions and predictions based on the data, as it provides insight into how much individual values might deviate from the expected outcome.
Z-score
A z-score is a statistical measurement that indicates how many standard deviations a data point is from the mean. It's an essential tool for comparing data points from different distributions or for finding the probability of a score occurring within a normal distribution.
To calculate a z-score, you use the formula:
\( z = \frac{X - \mu}{\sigma} \)
where \( X \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
  • In the exercise, the z-score was calculated for electricity usage of 1000 kWh. With a mean of 673 kWh and a standard deviation of 556 kWh, the z-score was \( 0.586 \).
  • This means the value of 1000 kWh is 0.586 standard deviations above the mean.
  • Z-scores help us place individual data points into context and find probabilities using the standard normal distribution table.
Z-scores are instrumental when assessing normal distribution, ensuring that the further a data point is from the mean, the less common it is within that distribution.
Cumulative Probability
Cumulative probability represents the probability that a randomly selected data point from a distribution will be less than or equal to a given value. In essence, it's a running total of probabilities, adding up all possibilities from the beginning of the distribution up to a specific value.
For example, if we are interested in the cumulative probability of households using up to 1000 kWh, we would look at the proportion of households that consume 1000 kWh or less from the total sample.
  • Using the standard normal distribution table, the cumulative probability associated with a z-score of 0.586 was found to be approximately 0.7224.
  • This indicates that 72.24% of households use 1000 kWh or less.
  • Cumulative probabilities allow us to assess where a particular data point lies in relation to the rest of the distribution, offering a clear understanding of relative standing within a dataset.
Cumulative probability is invaluable for interpreting statistical data as it offers insight into the likelihood of a score falling within a certain range, essential for decision making and risk assessment.

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Most popular questions from this chapter

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