6. According to the records of Enersource, an electric company serving the Mississauga area, the mean electricity consumption for all households during winter is 1650 kilowatt-hours per month. Assume that the monthly electricity consumption 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 has a monthly electricity consumption of 900 to 1300 kilowatt- hours?
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a. z = (x-mean)/ sd = (1950-1650)/320
To solve this problem, we need to use the concept of the normal distribution and the z-score.
a) To find the probability that the monthly electricity consumption is less than 1950 kilowatt-hours, we need to find the z-score and then find the corresponding cumulative probability.
1. Calculate the z-score:
The z-score formula is:
z = (x - μ) / σ
where x is the value we are interested in (1950 kilowatt-hours), μ is the mean (1650 kilowatt-hours), and σ is the standard deviation (320 kilowatt-hours).
z = (1950 - 1650) / 320
z = 300 / 320
z = 0.9375
2. Find the corresponding cumulative probability:
We need to find the area under the normal curve to the left of the z-score. We can use a table or use a calculator to find this probability. Let's use a calculator.
Using a calculator or software, we can find that the cumulative probability for a z-score of 0.9375 is approximately 0.8264.
So, the probability that the monthly electricity consumption during winter by a randomly selected household from this area is less than 1950 kilowatt-hours is approximately 0.8264.
b) To find the percentage of households with a monthly electricity consumption of 900 to 1300 kilowatt-hours, we need to find the z-scores for both values, calculate the cumulative probabilities, and then subtract the smaller cumulative probability from the larger cumulative probability.
1. Calculate the z-scores:
For 900 kilowatt-hours:
z1 = (900 - 1650) / 320
z1 = -3.28125
For 1300 kilowatt-hours:
z2 = (1300 - 1650) / 320
z2 = -1.09375
2. Find the corresponding cumulative probabilities:
Using a calculator or software, we can find that the cumulative probability for a z-score of -3.28125 is approximately 0.0006, and the cumulative probability for a z-score of -1.09375 is approximately 0.1368.
3. Subtract the smaller cumulative probability from the larger cumulative probability:
Percentage = (0.1368 - 0.0006) * 100
Percentage = 0.1362 * 100
Percentage = 13.62
So, approximately 13.62% of the households in this area have a monthly electricity consumption of 900 to 1300 kilowatt-hours.