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According to a government energy agency, the mean monthly household electricity bill in the United States in 2011 was $110.1. Assume the amounts are normally distributed with standard deviation $18.00. Use the TI-84 Plus calculator to answer the following.(a) What proportion of bills are greater than $131?

Question

According to a government energy agency, the mean monthly household electricity bill in the United States in 2011 was 110.1.Assumetheamountsarenormallydistributedwithstandarddeviation110.1. Assume the amounts are normally distributed with standard deviation 18.00. Use the TI-84 Plus calculator to answer the following.(a) What proportion of bills are greater than $131?

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Solution

To solve this problem, we will use the normal distribution function on the TI-84 Plus calculator. Here are the steps:

  1. First, calculate the z-score for 131.Thez−scoreiscalculatedas(X−μ)/σ,whereXisthevalueyou′reinterestedin,μisthemean,andσisthestandarddeviation.Inthiscase,X=131. The z-score is calculated as (X - μ) / σ, where X is the value you're interested in, μ is the mean, and σ is the standard deviation. In this case, X = 131, μ = 110.1,andσ=110.1, and σ = 18.00. So, the z-score is (131 - 110.1) / 18 = 1.16.

  2. Now, we will use the normalcdf function on the TI-84 Plus to find the proportion of bills greater than $131. To do this, press 2ND, then VARS to access the DISTR menu. Select 2:normalcdf.

  3. Enter the following values: lower bound = 1.16, upper bound = 1E99 (a very large number to represent infinity), μ = 0, σ = 1. This will calculate the area under the curve from the z-score to positive infinity.

  4. Press ENTER to calculate. The result is the proportion of bills that are greater than $131.

Remember, the result is a proportion, so if you want to express it as a percentage, you will need to multiply it by 100.

This problem has been solved

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