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According to records of Commonwealth Edison, monthly electric consumption for January follows a normal distribution with a mean of 1,650 kilowatt hours, and a standard deviation of 320 kilowatt hours. Commonwealth Edison wants to identify its heaviest users, which are those that are in the top 6% of electric consumption. How many kilowatt hours in January would qualify for the heavy user classification?

Respuesta :

Edufirst

Answer:

  • From 2,150 kilowatt hours will qualify for the heavy user classification.

Explanation:

Since the monthly electric consumption for January follows a normal distribution, you must use a standard normal table (Z table), which provides the probabilities for the (normalized) Z-scores.

The top 6% of electric consumption, means that you must find the z-score for which the probablitiy is ≥ 0.06.

The corresponding z-score shown in the table is 1.555.

Now just use the equation:

  • Z = ( x - μ) / σ

Where:

  • μ = mean = 1,650 kilowatt hours
  • σ = standard deviation = 320 kilowatt hours
  • x = the value of your variable, which you will determine

Substituting:

  • 1.555 = (x - 1,650) / 320
  • x = 320 × 1.555 + 1,650 = 2,148 ≈ 2,150
  • x ≈ 2,150 kilowatt hours.

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