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Chapter 6: Problem 45

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?

Short Answer

Expert verified
The probability that a randomly selected household consumes less than 1950 kilowatt-hours is the cumulative probability corresponding to the calculated Z score for Part a. The percentage of households that consume between 900 and 1300 kilowatt-hours is the difference between the probabilities corresponding to the calculated Z values for these limits, multiplied by 100.

Step by step solution

01

Calculate the Z score for part a

To solve part (a), first convert 1950 kilowatt-hours into a Z score. The formula to calculate Z score is \(Z = \frac{X - \mu}{\sigma}\), where X is the value from the dataset, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Substituting the given values, we get \(Z = \frac{1950 - 1650}{320}\).
02

Find the probability for part a

Once the Z-score has been calculated, use the Z-table to find the corresponding probability. A standard normal table shows the area under the curve to the left of any Z score, which translates into the cumulative probability. So, look up the Z-value obtained in a Z-table to get the required probability for part a.
03

Calculate the Z scores for part b

To solve part (b), convert 900 and 1300 kilowatt-hours into Z scores. Apply the Z score formula for each of these values separately.
04

Find the percentage for part b

Find the probabilities corresponding to both Z scores in a Z-table. The percentage of households that use between 900 and 1300 kilowatt-hours is the probability of the higher Z value minus the probability of the lower Z value. To convert this fractional probability into a percentage, multiply it by 100. This gives the final required percentage for part b.

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Most popular questions from this chapter

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