Question

Greeks bucked a global trend in which people in most countries expect their lives in five years to be better than their current lives. Even among those countries with much lower currrent life ratings, greater optimism was found because people cannot fathom their lives getting worse. But in Greece, only 25% expect their lives to be better in five years. In a simple random sample of 1100 Greek citizens, what is the probability that less than 27 percent of the sample expect their lives to be better in five years?

Answer 1

Answer:

The value is  $P( p < 0.27 ) = 0.67792$

Step-by-step explanation:

From the question we are told that

      The population proportion is  p= 0.25

       The sample size is  n  =  100

Generally given that the sample size is sufficiently large n > 30  , the mean of this sampling distribution is mathematically represented as

          $\mu_{x} = p = 0.25$

Generally the standard deviation is mathematically represented as

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

=>       $\sigma = \sqrt{ \frac{ 0.25 (1 - 0.25 )}{100} }$      

=>       $\sigma = 0.0433$

Generally the probability that less than 27 percent (0.27)of the sample expect their lives to be better in five year is mathematically represented as

      $P( p < 0.27 ) = P( \frac{ p - \mu_{x}}{\sigma} < \frac{0.27 - 0.25 }{ 0.0433} )$

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

=>     $P( p < 0.27 ) = P( Z< 0.4619 )$

From the z table the area under the normal curve to the left corresponding to 0.4619 is

     $P( Z< 0.4619 ) = 0.67792$

     $P( p < 0.27 ) = 0.67792$