Answer: a) 0.68, 0.1043, b) 0.68, 0.066
Explanation:
Since we have given that
p = 0.68
n= 20
Expected value = 0.68
So, Standard error would be

if n = 50
Expected value = 0.68
So, Standard error would be

In both the cases,
Yes it is appropriate to use the normal distribution for expected value and standard error.