Question

Given a population in which the probability of success is p = 0.20, if a sample of 500 items is taken, then: a) Calculate the probability that the proportion of successes in the sample will be between 0.18 and 0.23. b) Calculate the probability that the proportion of successes in the sample will be between 0.18 and 0.23 if the sample size is 200.

Answer 1

Solution

Given that,

p = 0.20

1 - p = 1-0.20

n = 500

= p =0.20

=  [p ( 1 - p ) / n] = [(0.20*0.80) / 500 ] = 0.0179

(A) P( 0.18 << 0.23)= P[(0.18 -0.20) /0.0179 < ( - ) / < (0.23-0.20) /0.0179 ]

= P(-1.12 < z <1.68 )

= P(z < 1.68) - P(z < -1.12)

Using z table

=0.9535-0.1314

=0.8221

probability= 0.8221

(B)

n=200

=  [p ( 1 - p ) / n] = [(0.20*0.80) / 200 ] = 0.0283

P( 0.18 << 0.23)= P[(0.18 -0.20) /0.0283< ( - ) / < (0.23-0.20) /0.0283 ]

= P(-0.71 < z <1.06 )

= P(z < 1.06) - P(z <- 0.71)

Using z table

=0.8554-0.2389

=0.6165

probability= 0.6165