Question

Suppose you take a simple random sample of size n=175 from the population of Reese's Pieces, still assuming that 45% of this population is orange (LaTeX: \piπ=.45).
What is the probability that a sample proportion of orange candies will be between .35 and .55?

Answer 1

Answer:

The  probability is  $P(0.35$

Step-by-step explanation:

From the question we are told that

  The sample size is  n =  175

   The population proportion is  p =  0.45

   

Generally the mean of the sampling distribution is  $\mu_{x} = 0.45$

 Generally the standard deviation is mathematically represented as  

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

=>      $\sigma_{x} = \sqrt{\frac{0.45 (1-0.45)}{175} }$

=>      $\sigma_{x} = 0.0376$

Generally the probability of  that the sample proportion of orange candies will be between 0.35 and  0.55 is  

      $P(0.35 < X < 0.55) = P( \frac{0.35 - 0.45}{0.0376} < \frac{X -\mu_{x}}{\sigma_{x}} < \frac{0.55 - 0.45}{0.0376} )$

=>   $P(0.35$

Generally $\frac{X - \mu_{x}}{\sigma_{x}} = Z (The \ standardized \ value \ of X)$

So

    $P(0.35$

=>  $P(0.35$

From the z-table  

    $P( Z< 2.6595 ) = 0.99609$

and

    $P( Z< - 2.6595 ) = 0.0039128$

So

   $P(0.35$

=> $P(0.35$