From a population of 200 elements, the standard deviation is known to be 14. a sample of 49 elements is selected. it is determined that the sample mean is 56. the standard error of the mean is

N = population size
n = sample size
sigma = population standard deviation
xbar = sample mean
SE = standard error
fpc = finite population correction

In this case,
N = 200
n = 49
sigma = 14
xbar = 56

Since n/N = 49/200 = 0.245 is larger than 0.05, this means we must use a finite population correction factor. I'll use fpc in place of 'finite population correction'.
If we ignore the fpc, then the SE would be simply sigma/sqrt(n) = 14/sqrt(49) = 2.
However we cannot ignore the fpc. We must use it due to the fact that n/N > 0.05. 



Let's compute the fpc factor
fpc = sqrt((N-n)/(N-1))
fpc = sqrt((200-49)/(200-1))
fpc = 0.87108780834612



With the fpc factor, we'll have the true SE to be SE = fpc*sigma/sqrt(n) = 0.87108780834612*14/sqrt(49) = 1.74217561669224

The final answer, accurate to 6 decimal places, is therefore 1.742176

There is no graph to go along with this