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DaniilM [7]
3 years ago
14

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.
Mathematics
1 answer:
Yuri [45]3 years ago
3 0

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

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