Respuesta :
Answer:
The confidence interval on this case is given by:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
For this case the confidence interval is given by (62.532, 76.478)[/tex]
And we can calculate the mean with this:
[tex] \bar X = \frac{62.532+76.478}{2}= 69.505[/tex]
So then the mean for this case is 69.505
Step-by-step explanation:
Previous concepts
The margin of error is the range of values below and above the sample statistic in a confidence interval.
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".
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
[tex]\bar X[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma[/tex] represent the population standard deviation
n represent the sample size
Assuming the X follows a normal distribution
[tex]X \sim N(\mu, \sigma)[/tex]
The confidence interval on this case is given by:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
For this case the confidence interval is given by (62.532, 76.478)[/tex]
And we can calculate the mean with this:
[tex] \bar X = \frac{62.532+76.478}{2}= 69.505[/tex]
So then the mean for this case is 69.505
