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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
a. The probability that the monthly electricity consumption during winter by a randomly selected household from this area is less than 1950 kilowatt-hours is found by calculating a Z-Score and looking up its corresponding probability in the standard normal distribution table. \n\nb. The percentage of the households in this area having a monthly electricity consumption of 900 to 1300 kilowatt-hours is obtained by finding the difference between two probabilities corresponding to two different Z-Scores.

Step by step solution

01

Find Z-Score for Part A

The Z-Score is a measure of how many standard deviations an element is from the mean. To calculate it, the formula is Z = (x - 渭) / 蟽. For this problem, x = 1950 kilowatt-hours, 渭 = 1650 kilowatt-hours, and 蟽 = 320 kilowatt-hours. Now, calculate the Z-Score using the given values.
02

Find Probability for Part A

Next, find the probability corresponding to the Z-Score calculated in the previous step. This can be done by looking up the Z-Score in the standard normal distribution table, which gives the area to the left of a given Z-Score. The area to the left of the Z-Score is the required probability.
03

Find Z-Scores for Part B

We have to find the percentage of households consuming between 900 to 1300 kilowatt-hours. Calculate two Z-Scores, one for \(X = 900\) kilowatt-hours and the other for \(X = 1300\) kilowatt-hours using the formula from Step 1.
04

Find Probability for Part B

Find the probabilities corresponding to both the Z-Scores calculated in the previous step. The difference between the two probabilities will give the percentage of households with electricity consumption between 900 to 1300 kilowatt-hours.

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