Question

In one region, the september energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kwh and a standard deviation of 218 kwh if 50 different homes are randomly selected, find the probability that their mean energy consumption level for september is greater than 1075 kwh.
your answer should be a decimal rounded to the fourth decimal place.

Answer 1

The probability that their mean energy consumption level for September is greater than 1075 kwh is $\boxed{0.2090}.$

Further explanation:

The Z score of the standard normal distribution can be obtained as,

$\boxed{{\text{Z}} = \dfrac{{X - \mu }}{{\dfrac{\sigma }{{\sqrt n }}}}}$

Here, Z is the standard normal value, $\mu$ represents the mean, $\sigma$ represents the standard deviation and n represents the observation

Given:

The mean of test is $\mu= 1050.$

The standard deviation of the waiting time is $\sigma=218.$

The number of observation is $n = 50.$

Explanation:

The probability that their mean energy consumption level for September is greater than 1075 can be calculated as follows,

$\begin{aligned}{\text{Probability}}&= P\left( {X > 1075} \right)\\&= P\left( {\dfrac{{X - \mu }}{{\dfrac{\sigma }{{\sqrt n }}}} > \dfrac{{1075 - \mu }}{{\dfrac{\sigma }{{\sqrt n }}}}} \right)\\&= P\left( {Z > \dfrac{{1075 - 1050}}{{\dfrac{{218}}{{\sqrt {50} }}}}} \right)\\&= P\left( {Z > \dfrac{{25}}{{30.83}}} \right)\\&= P\left( {Z > 0.81} \right)\\\end{aligned}$

The probability of $P\left( {Z > 0.81} \right)$ is $0.2090$ from the standard normal table or Z-table.

The probability that their mean energy consumption level for September is greater than $1075 kwh$ is $\boxed{0.2090}.$

Learn more:

1. Learn more about normal distribution brainly.com/question/12698949

2. Learn more about standard normal distribution brainly.com/question/13006989

3. Learn more about confidence interval of mean brainly.com/question/12986589

Answer details:

Grade: College

Subject: Statistics

Chapter: Confidence Interval

Keywords: one region, September, energy consumption, single family, homes, levels, mean, 1050 kwh, 218 kwh, standard normal distribution, standard deviation, test, measure, probability, low score, mean, repeating, indicated, normal distribution, percentile, percentage, proportion.

Answer 2

We first calculate the z-score corresponding to x = 1075 kWh. Given the mean of 1050 kWh, SD of 218 kWh, and sample size of n = 50, the formula for z is:
z = (x - mean) / (SD/sqrt(n)) = (1075 - 1050) / (218/sqrt(50)) = 0.81
From a z-table, the probability that z > 0.81 is 0.2090. Therefore, the probability that the mean of the 50 households is > 1075 kWh is 0.2090.