Complete Question
An analysis of several polls suggests that 60% of all Florida voters plan to vote for Anderson. A poll of 250 randomly selected Florida voters shows that 144 plan to vote for Anderson.
Required:
a. What is the probability of this result (i.e. 144 voters or less out of 250) happening by chance, assuming the aggregate poll model proportion of 60% is correct?
b. Does your result from part I indicate that the number of voters who plan to vote for Anderson has decreased? In other words, is this outcome unusual?(Recall that an unusual  event has a probability of 0.05 or less of occurring )
Answer:
a
[tex]P(p  <  \^p) =  0.2206[/tex]
b
It is not an unusual event
Step-by-step explanation:
From the question we are told that
  The population proportion is  p =  0.60
  The sample size is  n =  250
  The number that plans to vote for Anderson is  k =  144
Generally the mean of the sampling distribution is Â
    [tex]\mu _{x} =  p =  0.60[/tex]
Generally the standard deviation is Â
   [tex]\sigma  =  \sqrt{ \frac{p(1 - p )}{n} }[/tex]
=>   [tex]\sigma  =  \sqrt{ \frac{0.60(1 - 0.60 )}{250} }[/tex]
=>   [tex]\sigma  =  0.0310[/tex]
Generally the sample proportion is mathematically represented as Â
     [tex]\^{p} =  \frac{k}{n}[/tex]
=> Â Â Â [tex]\^{p} = Â \frac{144}{250}[/tex]
=> Â Â [tex]\^{p} = 0.576 [/tex]
Gnerally the probability of this result (i.e. 144 voters or less out of 250) happening by chance, assuming the aggregate poll model proportion of 60% is correct is mathematically represented as
 [tex]P(p  <  \^p) =  P(\frac{ p - \mu_{x}}{ \sigma }  <  \frac{ \^ p - \mu_{x}}{ \sigma } )[/tex]
=>   [tex]P(p  <  \^p) =  P(Z <  \frac{ 0.576 - 0,60}{ 0.0310} )[/tex]
=>    [tex]P(p  <  \^p) =  P(Z <  -0.77 )[/tex]
From the z-table  the probability of (Z <  -0.77)  is Â
   [tex]P(Z <  -0.77 )  =  0.2206[/tex]
So
  [tex]P(p  <  \^p) =  0.2206[/tex]
Since
 [tex]0.2206 > 0.05[/tex] it implies that it is not an unusual  event
