Respuesta :
Answer:
a) Standard error = 0.5
b) 99% Confidence interval: (19.2125,21.7875)
Step-by-step explanation:
We are given the following in the question:
Population mean = 20.50 mpg
Sample standard deviation = 3.00 mpg
Sample size , n = 36
Standard Error =
[tex]=\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = 0.5[/tex]
99% Confidence interval:
[tex]\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.575[/tex]
[tex]20.50 \pm 2.575(\displaystyle\frac{3}{\sqrt{36}})= 20.50 \pm 1.2875 = (19.2125,21.7875)[/tex]
The standard error and the 99% of the confidence interval is 0.5 and (19.2125, 21.7875)
What is normal a distribution?
It is also called the Gaussian Distribution. It is the most important continuous probability distribution. The curve looks like a bell, so it is also called a bell curve.
The highway fuel economy of a 2016 Lexus RX 350 FWD 6-cylinder 3.5-L automatic 5-speed using premium fuel is a normally distributed random variable with a mean of μ = 20.50 mpg and a standard deviation of σ = 3.00 mpg.
The sample size (n) is 36.
Then the standard error will be
[tex]\rm Standard \ error= \dfrac{\sigma}{\sqrt{n}}\\\\Standard \ error = \dfrac{3}{\sqrt{36}}\\\\Standard \ error = 0.5[/tex]
99% confidence interval
[tex]\rm \mu\ \pm z_{critical} \dfrac{\sigma }{\sqrt{n}}[/tex]
Put the value, we have
[tex]\rm z_{critical } \ at \ \alpha _{0.01} = \pm 2.575\\\\20.50 \pm 2.575 (\dfrac{3}{\sqrt{36}}) = 20.50 \pm 1.2875 = (19.2125, 21.7875)[/tex]
More about the normal distribution link is given below.
https://brainly.com/question/12421652
