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 …

Question

asked May 27, 2021 117k views

1 Answer

Vishnu Upadhyay · Answer 1 · 2021-06-01T12:32:39+0000

Answer:

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

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

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