If a confidence interval is given from 45.82 up to 55.90 and the mean is known to be 50.86, what is the margin of error?

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".

represent the sample mean  

population mean

s represent the sample standard deviation  

n represent the sample size  

Solution to the problem

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

  (1)

Or equivalently:

(2)

Where the margin of error is given by:

For this case we have the confidence interval limits given (45.82, 55.90)

We can find the width of the interval like this:

And now the margin of error would be the half of the width since we assume that the confidence interval is symmetrical.

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47