Please help me with this statistics question

Question

A survey showed that 76​% of adults need correction​ (eyeglasses, contacts,​ surgery, etc.) for their eyesight. If

19 adults are randomly​ selected, find the probability that no more than no more than 11 of them need correction for their eyesight. Is 11 a significantly low number of adults requiring eyesight​ correction?



Answer 1

Since 76% of the adults need some type of correction, then p=0.76.A sample is taken resulting in 11of 19 adults needing some form of correction. So p hat usually denoted by p with a carrot symbol above it is 0.58. Using the formula for a z-score z=(p hat - p)/sq root of p(1-p)/n. Therefore

z = (0.58 - 0.76)/sq root .(0.76)(0.24)/19 = -1.84

P(x less than or equal to 11) = 0.0329 or about 3.29%

What is considered unusual are results more than 2 standard deviations from the mean. The standard deviation is (0.76)(0.24)/19 = 0.098 0.76 - 2(0.098) = 0.564 which is less than p hat of 0.58. Therefore the sample proportion is within 2 standard deviations of the mean and considered not unusual.