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Electricity bills: According to a government energy agency, the mean monthly household electricity bill in the United States in 2011 was $109.77. Assume the amounts are normally distributed with standard deviation $22.00. Use the TI-84 Plus calculator to answer the following.(a) What proportion of bills are greater than $137?(b) What proportion of bills are between $85 and $148?(c) What is the probability that a randomly selected household had a monthly bill less than $119?Round the answers to at least four decimal places.Part 1 of 3The proportion of bills that are greater than $137 is .Part 2 of 3The proportion of bills that are between $85 and $148 is.Part 3 of 3The probability that a randomly selected household had a monthly bill less than $119 is .CheckSave For LaterSubmit Assignment

Question

Electricity bills: According to a government energy agency, the mean monthly household electricity bill in the United States in 2011 was 109.77.Assumetheamountsarenormallydistributedwithstandarddeviation109.77. Assume the amounts are normally distributed with standard deviation 22.00. Use the TI-84 Plus calculator to answer the following.(a) What proportion of bills are greater than 137?(b)Whatproportionofbillsarebetween137?(b) What proportion of bills are between 85 and 148?(c)Whatistheprobabilitythatarandomlyselectedhouseholdhadamonthlybilllessthan148?(c) What is the probability that a randomly selected household had a monthly bill less than 119?Round the answers to at least four decimal places.Part 1 of 3The proportion of bills that are greater than 137is.Part2of3Theproportionofbillsthatarebetween137 is .Part 2 of 3The proportion of bills that are between 85 and 148is.Part3of3Theprobabilitythatarandomlyselectedhouseholdhadamonthlybilllessthan148 is.Part 3 of 3The probability that a randomly selected household had a monthly bill less than 119 is .CheckSave For LaterSubmit Assignment

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Solution

To answer these questions, we will use the normal distribution function on the TI-84 Plus calculator.

(a) To find the proportion of bills that are greater than 137,weneedtocalculatethez−scorefor137, we need to calculate the z-score for 137. The z-score is calculated as (X - μ) / σ, where X is the value we're interested in, μ is the mean, and σ is the standard deviation. So, the z-score for $137 is (137 - 109.77) / 22 = 1.24. We then use the normalcdf function on the calculator to find the proportion of bills greater than this z-score, which is 1 - normalcdf(1.24, 1E99) = 0.1075.

(b) To find the proportion of bills that are between 85and85 and 148, we calculate the z-scores for these values and use the normalcdf function again. The z-score for 85is(85−109.77)/22=−1.13,andfor85 is (85 - 109.77) / 22 = -1.13, and for 148 it is (148 - 109.77) / 22 = 1.74. The proportion of bills between these values is normalcdf(-1.13, 1.74) = 0.7744.

(c) To find the probability that a randomly selected household had a monthly bill less than 119,wecalculatethez−scorefor119, we calculate the z-score for 119, which is (119 - 109.77) / 22 = 0.42. The probability is then normalcdf(-1E99, 0.42) = 0.6628.

So, the answers are:

Part 1 of 3: The proportion of bills that are greater than 137is0.1075.Part2of3:Theproportionofbillsthatarebetween137 is 0.1075. Part 2 of 3: The proportion of bills that are between 85 and 148is0.7744.Part3of3:Theprobabilitythatarandomlyselectedhouseholdhadamonthlybilllessthan148 is 0.7744. Part 3 of 3: The probability that a randomly selected household had a monthly bill less than 119 is 0.6628.

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