Question 993502
Looking at the problem, the expected number of complaints solved for 90 would be 90*0.65=58.5, and that is less than 62.  Therefore, the probability will be fairly large that they settle fewer than 62.

The probability is <62/90=0.6888 (repeat)
1 sample proportion
z=(0.6888-0.65)/sqrt{ (0.65)(0.35)/90}
denominator is 0.0503
z=  0.03888/0.05
We want the probability that z is < than that number,  The probability will be more than a half, since the sample proportion  is greater than the purported probability.
That is 0.7817