Question

What is the probability that a random sample of 400 U.S. youth will provide a sample proportion (p¯) that is within 0.03 of the population proportion (p)?

Answer 1

Complete Question

The proportion of adult women in the United States is 51% (p=0.51).What is the probability that a random sample of 400 U.S. youth will provide a sample proportion (p¯) that is within 0.03 of the population proportion (p)?

Answer:

The probability is  $P( 0.48 < p < 0.54 ) = 0.77$  or  $P( 0.48 < p < 0.54 ) = 77\%$

Step-by-step explanation:

From the question we are told that

   The sample size is  n  =  400

    The population proportion is  p = 0.51

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

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

=>      $\sigma = \sqrt{\frac{0.51(1-0.51)}{400} }$

=>      $\sigma = 0.02499$

Generally the lower limit for the range of the population proportion within 0.03 is

    $a = p - 0.03$

=>$a = 0.51 - 0.03$

=>$a = 0.48$

Generally the  upper  limit for the range of the population proportion within 0.03 is

    $b = p + 0.03$

=>$a = 0.51 + 0.03$

=>$a = 0.54$

Generally the  probability that a random sample of 400 U.S. youth will provide a sample proportion (p¯) that is within 0.03 of the population proportion (p) is mathematically represented as

     $P( a < p < b) = P( \frac{a - p }{\sigma } < \frac{\^ p - p }{\sigma} < \frac{b - p}{\sigma } )$

=>   $P( 0.48 < p < 0.54 ) = P( \frac{0.48 - 0.51 }{0.02499 } < \frac{\^ p - p }{\sigma} < \frac{0.54 - 0.51}{0.02499 } )$

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

=>   $P( 0.48 < p < 0.54 ) = P( -1.2 < Z < 1.2 )$

=>   $P( 0.48 < p < 0.54 ) = P( Z < 1.2) - P(Z < -1.2 )$

From the z table  the area under the normal curve to the left corresponding to  1.2 and  -1.2  is

     $P( Z < 1.2) =0.88493$

and  

    $P( Z < -1.2) =0.11507$

So

   $P( 0.48 < p < 0.54 ) = 0.88493 - 0.11507$

=>  $P( 0.48 < p < 0.54 ) = 0.77$