To solve this problem, I need to find the probability that the average weekly consumption over 12 weeks is less than 800 kWh.

Understanding the Problem

Given:

• Weekly electricity consumption is normally distributed
• Mean (μ) = 850 kWh
• Standard deviation (σ) = 95 kWh
• We need the probability that the average over 12 weeks is less than 800 kWh

Solution

When we take the average of multiple independent samples from a normal distribution, the sampling distribution of the mean follows these properties:

• The mean of the sampling distribution equals the population mean (μ)
• The standard error of the sampling distribution equals σ/√n, where n is the sample size

Let's calculate:

1. The mean of the sampling distribution = μ = 850 kWh
2. The standard error of the sampling distribution = σ/√n = 95/√12 = 95/3.464 = 27.425 kWh
3. To find the probability that the average is less than 800 kWh, I need to calculate the z-score:

4. Now I need to find P(Z < -1.823) using the normal distribution table.

For negative z-values, we use Φ(-z) = 1 - Φ(z)

Looking at the table for z = 1.82 (closest to 1.823): Φ(1.82) = 0.9656

Therefore: P(Z < -1.823) = 1 - Φ(1.823) ≈ 1 - 0.9656 = 0.0344

To get a more precise value, I'll interpolate between z = 1.82 and z = 1.83: Φ(1.82) = 0.9656 Φ(1.83) = 0.9664 Φ(1.823) ≈ 0.9656 + (0.9664 - 0.9656) × (0.823 - 0.82)/(0.83 - 0.82) ≈ 0.9658

Therefore: P(Z < -1.823) = 1 - 0.9658 = 0.0342

The probability that the average weekly consumption over 12 weeks is less than 800 kWh is 0.034 (rounded to 3 decimal places).

Mathematical Verification

The Central Limit Theorem tells us that when we take samples of size n from a population with mean μ and standard deviation σ, the sampling distribution of the sample mean will be approximately normal with:

• Mean = μ
• Standard deviation = σ/√n

In this case:

• Population mean (μ) = 850 kWh
• Population standard deviation (σ) = 95 kWh
• Sample size (n) = 12 weeks

Therefore, the sampling distribution of the mean has:

• Mean = 850 kWh
• Standard deviation = 95/√12 = 27.425 kWh

The probability that the sample mean is less than 800 kWh is: P(X̄ < 800) = P(Z < (800-850)/27.425) = P(Z < -1.823) = 0.034

The answer is 0.034.

To be precise, the probability that the average weekly consumption over 12 weeks is less than 800 kWh is 0.034 (rounded to 3 decimal places).