Question

In one region, the september energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kwh and a standard deviation of 218 kwh. for a randomly selected home, find the probability that the september energy consumption level is between 1100 kwh and 1225 kwh.
your answer should be a decimal rounded to the fourth decimal place.

Answer 1

The probability would be 0.1971.

We will calculate a z-score for each end of this interval.

z = (X-μ)/σ

For the lower limit:
z = (1100-1050)/218 = 50/218 = 0.23

For the upper limit:
z = (1225-1050)/218 = 175/218 = 0.80

Using a z-table (http://www.z-table.com) we see that the area under the curve to the left of, less than, the lower limit is 0.5910.  The area under the curve to the left of, less than, the upper limit is 0.7881.  To find the area between them, we subtract:

0.7881 - 0.5910 = 0.1971

Answer 2

Answer: FULL ANSWER TEST ON MY QUIZLET LINK: Lesson 6: Unit Test Statistics A Unit 8: Normal Probability Distributions.. sadly missing one question I forgot to include but 100% guaranteed.. READ QUESTIONS CAREFULLY

Step-by-step explanation:

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