Question

A polling agency is investigating the voter support for a ballot measure in an upcoming city election. The agency will select a random sample of 500 voters from one region, Region A, of the city. Assume that the population proportion of voters who would support the ballot measure in Region A is 0.47. What is the probability that the proportion of voters in the sample of Region A who support the ballot measure is greater than 0.50

Answer 1

Answer:

The value is  $P( X > 0.50) = 0.089264$

Step-by-step explanation:

From the question we are told that

   The sample size is  n =  500

   The  population proportion is  p =  0.47  

     

Generally given that the sample size is sufficiently large , the mean of this sampling distribution is mathematically represented as

        $\mu_x = p = 0.47$

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

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

=>     $\sigma = \sqrt{\frac{ 0.47 (1-0.47 )}{ 500 } }$

=>     $\sigma = 0.0223$

Gnerally the probability that the proportion of voters in the sample of Region A who support the ballot measure is greater than 0.50 is mathematically represented as

          $P( X > 0.50) = P( \frac{ X - \mu }{ \sigma } > \frac{ 0.50 - 0.47 }{ 0.0223 } )$

$\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )$

=>        $P( X > 0.50) = P( Z> 1.3453 )$

From the z table  the area under the normal curve to the left corresponding to  1.3453   is

          $P( Z> 1.3453 ) = 0.089264$

So  

          $P( X > 0.50) = 0.089264$