Question

In one region of alberta, the september energy consumption levels for single-family homes had a mean of 1050 kwh and a standard deviation of 218 kwh. a. find the probability that a randomly selected single-family home has an energy consumption level greater than 1075 kwh for september.

Answer 1

To find the probability that a randomly selected single-family home has an energy consumption level greater than 1075 kwh, calculate the z-score and use the standard normal distribution table to find the corresponding probability.

Step-by-step explanation:

To find the probability that a randomly selected single-family home has an energy consumption level greater than 1075 kwh for September, we need to calculate the z-score and then find the corresponding probability using the standard normal distribution table.

The z-score can be calculated using the formula: z = (x - mean) / standard deviation. In this case, the mean is 1050 kwh and the standard deviation is 218 kwh. So, z = (1075 - 1050) / 218 = 25 / 218 = 0.1147.

Now, we can use the standard normal distribution table to find the probability associated with a z-score of 0.1147. The probability is approximately 0.5478, or 54.78%.