Question

According to the Commerce Department in the Spring of 2016, 65% of U.S. households had some form of high-speed internet connection. Suppose 35 U.S. households were selected at random, find the probability that:

1. Exactly 15 of these households had a high-speed internet connection…………………………….

2. Between 16 and 25 had (inclusive) had a high-speed internet connection.........................

3. Less than 18 had a high-speed internet connection…………………………………......

4. More than 28 had a high-speed internet connection

Answer 1

X ~ Bin ( n , p)

Where n = 35 , p = 0.65

Mean = n p = 35 * 0.65 = 22.75

Standard deviation = sqrt [ np ( 1 - p) ] = sqrt [ 35 * 0.65 ( 1 - 0.65) ] = 2.8218

Using normal approximation

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

a)

With continuity correction

P(X = 15) = P(14.5 < X < 15.5)

P ( 14.5 < X < 15.5 ) = P ( Z < ( 15.5 - 22.75 ) / 2.8218 ) - P ( Z < ( 14.5 - 22.75 ) / 2.8218 )
= P ( Z < -2.57) - P ( Z < -2.92 )
= 0.0051 - 0.0018
= 0.0033
b)

With continuity correction

P(16 <= X <= 25) = P(15.5 < X < 25.5)

P ( 15.5 < X < 25.5 ) = P ( Z < ( 25.5 - 22.75 ) / 2.8218 ) - P ( Z < ( 15.5 - 22.75 ) / 2.8218 )
= P ( Z < 0.97) - P ( Z < -2.57 )
= 0.834 - 0.0051
= 0.8289

c)

With continuity correction

P(X < 18) = P(X < 17.5)

P ( ( X < 17.5 ) = P ( Z < 17.5 - 22.75 ) / 2.8218 )
= P ( Z < -1.86 )
P ( X < 17.5 ) = 0.0314

d)

With continuity correction

P(X > 28) = P(X >28.5 )  

P ( X > 28.5 ) = P(Z > (28.5 - 22.75 ) / 2.8218 )
= P ( Z > 2.04 )
= 1 - P ( Z < 2.04 )
= 1 - 0.9793

= 0.0207