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. Find P45, which is the consumption level separating the bottom 45% from the top 55%

Answer 1

Answer:

$a=1050 -0.126*218=1022.532$

So the value of height that separates the bottom 45% of data from the top 55% is 1022.532.  

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Solution to the problem

Let X the random variable that represent the consumption levels of a population, and for this case we know the distribution for X is given by:

$X \sim N(1050,218)$  

Where $\mu=1050 kWh$ and $\sigma=218kWh$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.55$   (a)

$P(X<a)=0.45$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.45 of the area on the left and 0.55 of the area on the right it's z=-0.26. On this case P(Z<-0.126)=0.45 and P(z>-0.126)=0.55

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.45$  

$P(z<\frac{a-\mu}{\sigma})=0.45$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.126=\frac{a-1050}{218}$

And if we solve for a we got

$a=1050 -0.126*218=1022.532$

So the value of height that separates the bottom 45% of data from the top 55% is 1022.532.  

Answer 2

Using the normal distribution, it is found that the 45th percentile consumption is of 1023 kWh.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• Mean of 1050kWh, hence $\mu = 1050$.

• Standard deviation of 218 kWh, hence $\sigma = 218$.

The 45th percentile is X when Z has a p-value of 0.45, so X when Z = -0.125.

$Z = \frac{X - \mu}{\sigma}$

$-0.125 = \frac{X - 1050}{218}$

$X - 1050 = -0.125(218)$

$X = 1023$

Hence, the 45th percentile consumption is of 1023 kWh.

To learn more about the normal distribution, you can take a look at https://brainly.com/question/24663213