Question

suppose that 17% of unmarried couples are mixed racially.

a) when n=150, what is the probability that the proportion of mixed couples will be greater than 0.2?

b) when n=200, what is probability that the proportion of mixed couples will be greater than 0.2?

Answer 1

Solution

Given that,

p = 0.17

1 - p = 1-0.17=0.83

n = 150

= p =0.17

=  [p( 1 - p ) / n] = [(0.17*0.83) / 150 ] = 0.0307

P( >0.2 ) = 1 - P( <0.2 )

= 1 - P(( - ) / < (0.2-0.17) / 0.0307)

= 1 - P(z <0.98 )

Using z table

= 1 -0.8365

=0.1635

probability=0.1635

(B)

n = 200

= p =0.17

=  [p( 1 - p ) / n] = [(0.17*0.83) / 200 ] = 0.0266

P( >0.2 ) = 1 - P( <0.2 )

= 1 - P(( - ) / < (0.2-0.17) / 0.0266)

= 1 - P(z <1.13 )

Using z table

= 1 -0.8708

=0.1292

probability=0.1292