Answer:
CI 90 %  =  [  0,165  ;  0,235 ]
Lower limit     0,165
upper limit      0,235
Step-by-step explanation:
Sample information:
sample size    n  =  225
number of dissatisfied  individuals   x  =  45
p = 45/225
p = 0,2    and  q =  1 -  p   q  =  1  -  0,2   q  =  0,8
p*n = 0,2*225   =  45    and   q*n   =  0,8*225  =  180
p*n  and q*n  big enough to use the approximation of binomial distribution to normal distribution
90 % of Confidence Interval   then a significance level is  α  = 10%
α  =  0,1 in z table we get z(c) for that significance level
z(c) = 1,28
CI 90 %   =  [  p  ±  z(c)*√(p*q)/n ]
z(c) * √(p*q)/n    =  1,28 *  √ ( 0,2*0,8)/225
z(c) * √(p*q)/n    =  1,28 * 0,027
z(c) * √(p*q)/n    =  0,035
CI 90 %  =  [  0,2  ±  0,035 ]
CI 90 %  =  [  0,165  ;  0,235 ]
Lower limit     0,165
upper limit      0,235