Question

. The CBS television show 60 Minutes recently had a (market) share of 20, meaning that 20% of TV sets in use were tuned to 60 Minutes during its airing hour. An advertiser wants to conduct its own survey of 200 households using the 20% share value. Using the normal approximation to the binomial, with X=number of TV sets tuned to 60 Minutes at the appropriate hour, compute (a) P(X equal to or less than 35), (b) P(X is between 35 and 45 inclusive), (c) P(X equal to or greater than 50). Don’t forget the continuity correction.

Answer 1

n = 200

p = 0.2

µ = n * p = 200 * 0.2 = 40

sd = sqrt(n * p * (1 - p)) = sqrt(200 * 0.2 * 0.8) = 5.657

a) P(X < 35) = P(X < 35.5)

                   

                   

                    = P(Z < -0.8)

                    = 0.2119

b) P(35 < X < 45) = P(34.5 < X < 45.5)

                           

                           

                            = P(-0.97 < Z < 0.97)

                            = P(Z < 0.97) - P(Z < -0.97)

                            = 0.8340 - 0.1660

                            = 0.6680

c) P(X > 50) = P(X > 49.5)

                   

                   

                    = P(Z > 1.68)

                    = 1 - P(Z < 1.68)

                    = 1 - 0.9535

                    = 0.0465