Question

Of the total population of the United States, 20% live in the northeast. If 2000 residents of the United States are selected at random, approximate the probability that at least 50 live in the northeast.

Answer 1

X ~ Bin ( n , p)

Where n = 2000 , p = 0.20

Mean = n p = 2000 * 0.20 = 400

Stadnard deviation = sqrt [ n p ( 1 - p) ] = sqrt [ 2000 * 0.20 ( 1 - 0.20) ] = 17.8885

Using normal approximation,

P(X < x) = P(Z < ( x - mean) / SD )

P ( X > 49.5 ) = P(Z > (49.5 - 400 ) / 17.8885 )
= P ( Z > -19.59 )
= 1 - P ( Z < -19.59 )
= 1 - 0 (From Z table)

= 1