Question

Suppose 49% of the population has a retirement account. If a random sample of size 896 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%?

Answer 1

The probability is
`P( |\^ p - p | < 0.03 ) = 0.9275`

Explanation:

From the question we are told that

The population proportion is
`p = 0.49`

The sample size is n = 896

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

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

=>
`\sigma_(p) = \sqrt{ (0.49(1 -0.49 ))/(896) }`

=>
`\sigma_(p) = 0.0167`

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

`P( |\^ p - p | < 0.03 ) = P( ( | \^ p - p|)/(\sigma_p ) < (0.03)/(0.0167) )`

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

=>
`P( |\^ p - p | < 0.03 ) = P(|Z| < 1.796 )`

=>
`P( |\^ p - p | < 0.03 ) = P(- 1.796 < Z < 1.796 )`

=>
`P( |\^ p - p | < 0.03 ) = P(1.796 < Z) - P(Z < -1.796 )`

From the z table the area under the normal curve to the left corresponding to 1.796 and - 1.796 is

`P(Z < -1.796 ) = 0.036247`

`P(1.796 < Z) = 0.96375`

So

`P( |\^ p - p | < 0.03 ) = 0.96375 - 0.036247`

=>
`P( |\^ p - p | < 0.03 ) = 0.9275`