Question

Suppose a large shipment of compact discs contained 19% defectives. If a sample of size 343 is selected, what is the probability that the sample proportion will differ from the population proportion by less than 4%? Round your answer to four decimal places.

Answer 1

The value is
`P( | \^p - p | < 0.04) = 0.9408`

Explanation:

From the question we are told that

The population proportion is
`p = 0.19`

The sample size is n = 343

Generally given that the ample size is large enough , i.e n > 30 then the mean of this sampling distribution is mathematically represent

`\mu_(x) = p = 0.19`

Generally the standard deviation is mathematically represented as

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

=>
`\sigma =\sqrt{(0.19 (1- 0.19 ))/(343 ) }`

=>
`\sigma = 0.0212`

Generally the the probability that the sample proportion will differ from the population proportion by less than 4% is mathematically represented as

`P( | \^p - p | < 0.04) = P( (|\^ p - p |)/( \sigma_p ) < (0.04)/(0.0212 ) )`

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

`P( | \^p - p | < 0.04) = P( |Z| < 1.887 )`

=>
`P( | \^p - p | < 0.04) = P( Z < 1.887 )- P( Z < -1.887 )`

From the z table the area under the normal curve to the left corresponding to 1.887 and - 1.887 is

`P( Z < 1.887 )= 0.97042`

and

`P( Z < -1.887 )= 0.02958`

So

`P( | \^p - p | < 0.04) = 0.97042 - 0.02958`

=>
`P( | \^p - p | < 0.04) = 0.9408`