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

Answer:

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

Step-by-step 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

      $\sigma =\sqrt{\frac{p(1- p)}{n} }$

=>   $\sigma =\sqrt{\frac{0.19 (1- 0.19 )}{343 } }$  

=>   $\sigma = 0.0212$

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

        $P( | \^p - p | < 0.04) = P( \frac{|\^ p - p |}{ \sigma_p } < \frac{0.04}{0.0212 } )$

$\frac{|\^ p - p |}{\sigma } = |Z| (The \ standardized \ value\ of \ |\^ p - p | )$

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

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

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

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

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