Question

Suppose 48% of the population has a retirement account. If a random sample of size 632 is selected, what is the probability that the proportion of persons with a retirement account will differ from the population proportion by greater than 3%? Round your answer to four decimal places.

Answer 1

Population proportion, p = 0.48

Sample size, n = 632

P( < A) = P(Z < (A - )/)

= p = 0.48

=

=

= 0.0199

P(a retirement account will differ from the population proportion by greater than 3%) = P( < 0.48-0.03) + P( > 0.48 + 0.03)

= P( < 0.45) - P( > 0.51)

= P( < 0.45) - [1 - P( < 0.51)]

= P(Z < (0.45 - 0.48)/0.0199) - 1 + P( < 0.51 - 0.48)/0.0199)

= P(Z < -1.51) - 1 + P(Z < 1.51)

= 0.0655 + 1 - 0.9345

= 0.1310