# 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?

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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**

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