Answer:
The likelihood is [tex]P(p < \^ p) = 0.066807[/tex]
Step-by-step explanation:
From the question we are told that
The population proportion is p =0.55
The sample size is n = 619
The sample proportion is [tex]\^ p = 0.52[/tex]
Generally the mean of the sampling distribution is [tex]\mu_{x} = p = 0.55[/tex]
The standard deviation is [tex]\sigma = \sqrt{\frac{p(1- p )}{n} }[/tex]
=> [tex]\sigma = \sqrt{\frac{0.55(1- 0.55 )}{619} }[/tex]
=> [tex]\sigma = 0.02[/tex]
Generally the likelihood that the administration have obtained a sample proportion as low as or lower than it obtained from its survey is mathematically represented as
[tex]P(p < \^ p) = P(\frac{p- \mu}{\sigma } < \frac{ 0.52- 0.55}{0.02 } )[/tex]
Generally [tex]\frac{p- \mu}{\sigma } = Z(The \ standardized \ value \ of \ p)[/tex]
=> [tex]P(p < \^ p) = P(Z <-1.5 )[/tex]
From the z-table the probability of (Z <-1.5 ) is
[tex]P(Z <-1.5 ) = 0.066807[/tex]
So
[tex]P(p < \^ p) = 0.066807[/tex]
