Question 1103717
p represents the probability that customers are satisfied, therefore p = .9.
q represents the probability that customers are not satisfied, therefore q = 1 - .9 = .1
n = 120 which is your sample size.


the mean of this sample is equal to n * p which is equal to 120 *.9 = 108.
the standard deviation of this sample is equal to sqrt(n*p*q) which is equal to sqrt(120*.9*.1) = 3.286335345.


the mean of this sample is equal to 108.
the standard deviation of this sample is equal to 3.286335345.


you are looking to find the probability of getting a score of less than 100.


the z-score is given by the formula of z = (x-m)/s


x is the test score.
m is the mean.
s is the standard error of the sample.


formula becomes z = (100-108)/3.286335345.


solve for z to get z = -2.434322478


the critical z-score for a 99% confidence interval is equal to plus or minus 2.575829303.


the critical z-score for a 95% confidence interval is equal to plus or minus 1.959963986.


your sample is borderline within limits at 99% confidence interval, and clearly out of limits at 95% confidence interval.


at 99% confidence interval, with a sample size of 120, the number of satisfied customers should not be less than 108 - 2.575829303 * 3.286335345 = 99.53496112.


at 95% confidence interval, with a sample size of 120, the number of satisfied customers should not be less than 108 - 1.959963986 * 3.286335345 = 101.5589011.


either way, the probability is unlikely that getting less than or equal to 100 satisfied customers out of a sample of 120 customers is due to random variations in sample means only.


accordingly, i believe that this constitutes sufficient evidence that the precentage of satisfied customer is probably less than 90%.