Question

Suppose 55U% of the population has a retirement account. If a random sample of size 639639 is selected, what is the probability that the proportion of persons with a retirement account will differ from the population proportion by less than 3%3%

Answer 1

The probability is
`P(|\^ p - p | < 0.03) = 0.87224`

Explanation:

From the question we are told that

The population proportion is p = 0.55

The sample size is n = 639

Generally the standard deviation of this sampling distribution is mathematically represented as

`\sigma = \sqrt{(p(1 -p))/(n) }`

=>
`\sigma = \sqrt{(0.55 ( 1 -0.55))/(639 ) }`

=>
`\sigma = 0.0197`

Generally the probability that the proportion of persons with a retirement account will differ from the population proportion by less than 3% is mathematically evaluated as

`P(|\^ p - p | < 0.03) = P( (|\^ p - p |)/(\sigma ) < (0.03)/(0.0197) )`

`(|\^ p-p|)/(\sigma )  =  |Z| (The  \ standardized \  value\  of  \ |\^ p - p|)`

`P(|\^ p - p | < 0.03) = P( {|Z| < 1.523 )`

=>
`P(|\^ p - p | < 0.03) = P( -1.523 \le Z \le 1.523 )`

=>
`P(|\^ p - p | < 0.03) = P( Z \le 1.523) - P( Z \le -1.523 )`

From the z table the area under the normal curve to the left corresponding to -1.523 is

`P(Z \le -1.523 ) = 0.063879`

From the z table the area under the normal curve to the left corresponding to 1.523 is

`P(Z \le 1.523 ) = 0.93612`

`P(|\^ p - p | < 0.03) = 0.93612 - 0.063879`

`P(|\^ p - p | < 0.03) = 0.87224`