Question

This Question: 1 pt number of aduts requiring eyesight conrection The probability that at least 10 of the 11 adults require eyesight correction is (Round to three decimal places as needed) s 10 a significanty high number of aduits requring eyesight conection? Note that a small probability is one that is less han 0 05 O C. Yes, becauan the probablity of this occurring is not smal Click to select your answer

Answer 1

Let X denote the number of people need correction for their eyesight Then X follows a Binomial distribution with n = 11 and p = 0.81. therefore P(X = x) = n!/x!(n - x)! p^x (1 - p)^n - x = 11!/x!(11 - x)! (0.81)^x (0.19)^11 - x, (x = 0, 1, 2, 3, . . . 11) P(X greater than or equal to 10) = sigma^11_x = 10 P(X = x) = sigma^11_x = 10 11!/x!(11 - x)! (0.81)^x (0.19)^11 - x option C