Question

Suppose the number of inches of rainfall each year in a city is normally distributed. For a random sample of years, the confidence interval (3.9,7.7) is generated. Find the margin of error

Answer 1

Answer:

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Or equivalently:

$\bar X \pm ME$

For this case we have the interval given (3.9, 7.7) and we want to find the margin of error. Using the property of symmetry for a confidence interval we can estimate the margin of error with this formula:

$ME= \frac{Upper -Lower}{2}= \frac{7.7-3.9}{2}= 1.9$

Step-by-step explanation:

Previous concepts

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".

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".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Or equivalently:

$\bar X \pm ME$

For this case we have the interval given (3.9, 7.7) and we want to find the margin of error. Using the property of symmetry for a confidence interval we can estimate the margin of error with this formula:

$ME= \frac{Upper -Lower}{2}= \frac{7.7-3.9}{2}= 1.9$