Question

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?

Answer 1

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

Step-by-step 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.