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According to the records of an electric company serving the Boston area, the mean electricity consumption during winter for all households is 1650 kilowatt-hours per month. Assume that the monthly electric consumptions during winter by all households in this area have a normal distribution with a mean of 1650 kilowatt-hours and a standard deviation of 320 kilowatt-hours. The company sent a notice to Bill Johnson informing him that about \(90 \%\) of the households use less electricity per month than he does. What is Bill Johnson's monthly electricity consumption?

Short Answer

Expert verified
Bill Johnson's monthly electricity consumption is approximately 2064 kilowatt-hours.

Step by step solution

01

Identify the necessary parameters

The mean (\(\mu\)) of the normal distribution is given as 1650 kWh, and the standard deviation (\(\sigma\)) is given as 320 kWh. Additionally, we know that Bill is in the \(90th\) percentile, which means he uses more electricity than \(90\%\) of the other households. This percentile corresponds to a z-score in a standard normal distribution.
02

Find the corresponding z-score

A z-score is a measure of how many standard deviations an element is from the mean. The z-score of \(90\%\) of a standard normal distribution can be found in the standard normal distribution table or calculated by a calculator. Using either of these methods gives a z-score of approximately \(1.28\).
03

Calculate the monthly electricity consumption

To find Bill's electricity consumption, we use the formula for transforming a standard z-score to a score on the original measurement scale, which is \(X = \mu + Z\sigma\), where \(X\) is the measurement, \(\mu\) is the mean, \(Z\) is the z-score, and \(\sigma\) is the standard deviation. Substituting the values, we have \(X = 1650 + 1.28*320\). This gives \(X \approx 2064\) kWh.

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Key Concepts

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

Mean and Standard Deviation
In statistics, the concepts of mean and standard deviation are fundamental when discussing data sets. The mean, or average, is a measure that represents the central point of a dataset. It is calculated by summing all data points and dividing by the number of points. For example, in our electricity consumption scenario, the mean consumption for households is 1650 kilowatt-hours per month.

On the other hand, the standard deviation provides insight into how spread out the data points are around the mean. A low standard deviation indicates that the data points tend to be close to the mean, whereas a high standard deviation suggests they are more spread out. In this case, a standard deviation of 320 kilowatt-hours means that individual household electricity consumption can vary significantly from the mean of 1650 kilowatt-hours. This variability is crucial for understanding the distribution of electricity usage among different households.
Percentile and Z-Score
Percentiles are a way of understanding the relative position of a data point within a dataset. If a score is in the 90th percentile, it means that it exceeds 90% of the other scores. In the context of our example, Bill's electricity consumption is higher than 90% of the households, indicating he is at the 90th percentile.

The z-score, on the other hand, is a statistical measure that tells us how many standard deviations a particular score is from the mean. It is crucial in transforming a data point from a standard normal distribution to a specific dataset. For the 90th percentile, the z-score is approximately 1.28. This number helps us translate the percentile rank into a real-life measurement using the formula:
  • \(X = \mu + Z\sigma\)
  • \(\mu\) is the mean
  • \(Z\) is the z-score
  • \(\sigma\) is the standard deviation
By applying these values, we calculate Bill's electricity consumption.
Electricity Consumption Analysis
Analyzing electricity consumption, especially during specific seasons like winter, is essential for energy companies and households alike. Such analyses help in understanding usage patterns and preparing adequate resources to meet demand.

In our scenario, the electric company has provided that the typical household consumes on average 1650 kilowatt-hours per month. Knowing these details helps the company predict overall energy needs and anticipate any potential issues in supply or overly high consumption periods. Understanding these elements is not only beneficial for the company but also for consumers. It allows households like Bill's to compare their usage against typical consumption rates and adjust their habits if necessary.

Using statistics such as mean, standard deviation, percentiles, and z-scores, households and companies can make informed decisions. This means better energy efficiency and resource management across the board.

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Most popular questions from this chapter

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According to the records of an electric company serving the Boston area, the mean electricity consumption for all households during winter is 1650 kilowatt-hours per month. Assume that the monthly electricity consumptions during winter by all households in this area have a normal distribution with a mean of 1650 kilowatt-hours and a standard deviation of 320 kilowatt-hours. a. Find the probability that the monthly electricity consumption during winter by a randomly selected household from this area is less than 1950 kilowatt- hours. b. What percentage of the households in this area have a monthly electricity consumption of 900 to 1300 kilowatt-hours?

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