Question

According to the records of an electric company serving the Boston area, the mean electricity consumption during winter for all households is 1650 kilowatt-hours per month. Assume that the monthly electricity consumption during winter by all households in this area have a normal distribution with a mean of 1650 kilowatt-hours and a standard deviation of 320 kilowatt-hours. The company sent a notice to Bill Johnson informing him that about 75% of the households use less electricity per month than he does. What is Bill Johnson’s monthly electricity consumption?Round your answer to the nearest integer.Bill Johnson’s monthly electricity consumption is approximately Enter your answer in accordance to the question statement kWh.

Answer 1

Answer:

Bill's monthly consumption is 1866 kWh

Step-by-step explanation:

Given:

mean = 1650 kWh

standard deviation = 320kWh

% of those that use less electricity = 75%

To find:

Bill Johnson's monthly consumption

To find the monthly consumption of Bill, we will use the z-score formula:

`\begin{gathered} z=(x-μ)/(σ) \\ where\text{ }μ\text{ = mean} \\ σ\text{ = standard deviation} \\ x\text{ = ?} \\ z\text{ = z score} \end{gathered}`
`\begin{gathered} z\text{ = }\frac{x-\text{ 1650}}{320} \\ 75th\text{ percentile of z score = 0.675} \\ z\text{ = 0.675} \\ \\ 0.675=\frac{x-\text{1650}}{320} \end{gathered}`
`\begin{gathered} simplify: \\ 0.675(320)\text{ = x - 1650} \\ 216\text{ = x - 1650} \\ x\text{ = 216 + 1650} \\ x\text{ = 1866 kWh} \end{gathered}`

Hence, Bill's monthly consumption is 1866 kWh