Question

According to an article in the American Demographics, 19% of the entire U.S. population listens to Internet radio (Rebecca Gardyn, “High Frequency,” July 2000, 32-36). Suppose a random sample of 100 U.S. citizens is selected. The probability that the sample proportion of U.S. citizens listening to Internet radio is more than 0.25 is.

Answer 1

Solution:

Given ,

p = 0.19 = 0.19  (population proportion)

1 - p = 1 - 0.19 = 0.81

n = 100 (sample size)

Let be the sample proportion.

The sampling distribution of is approximately normal with

mean = =  p = 0.19

SD =   =     

=  \sqrt{0.19(1-0.19)/100}

=  0.03923009049

Find P(Sample proportion is more than 0.25)

= P( > 0.25)

=  

=  P(Z > (0.25 -0.19)/0.03923009049)

= P(Z > 1.53)

= 1 - P(Z < 1.53)

= 1 - 0.9370   ...use z table

= 0.0630

P(Sample proportion is more than 0.25 ) = 0.0630