It is estimated that 62% of television viewers channel surf during commercials. A market research from 1500 television viewers. What is the probability that at least 940 of them were channel surfers?
The estimated percentage of television viewers that channel surf during commercials is 62%. A market research from 1500 television viewers is conducted. The probability of at least 940 of them being channel surfers has to be determined.
The total number of viewers surveyed is n = 1500. The probability of a viewer surfing channels is p = 0.62; n*p = 930. The probability of a viewer not surfing channels is q = (1 - 0.62) = 0.38; n*q = 570. As both n*q and n*p are greater than 5 it is possible to calculate an approximate value for mean and standard deviation.
mean `mu = 1500*0.62 = 930` and the standard deviation `sigma = sqrt(n*p*q) = sqrt(1500*0.38*0.62)` = `sqrt 353.4 ~~ 18.79`
Using the values derived the z-score is `(940 - 930)/18.79 = 10/18.97 = 0.5319` . Using a z-score table the associated probability is 0.7019
The probability that at least 940 of the viewers surf channels is 0.7019