Mathematics College

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?

Answers

Answer 1

Answer:

The  probability is  [tex]P(0.35 <X < 0.55) = 0.9921772[/tex]

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  [tex]\mu_{x} = 0.45[/tex]

 Generally the standard deviation is mathematically represented as  

        [tex]\sigma_{x} = \sqrt{\frac{p (1-p)}{n} }[/tex]

=>      [tex]\sigma_{x} = \sqrt{\frac{0.45 (1-0.45)}{175} }[/tex]

=>      [tex]\sigma_{x} = 0.0376[/tex]

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

      [tex]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} )[/tex]

=>   [tex]P(0.35 <X < 0.55) = P( -2.696 < \frac{X -\mu_{x}}{\sigma_{x}} < 2.6595 )[/tex]

Generally [tex]\frac{X - \mu_{x}}{\sigma_{x}} = Z (The \ standardized \ value \ of X)[/tex]

So

    [tex]P(0.35 <X < 0.55) = P( -2.6596 < Z< 2.6595 )[/tex]

=>  [tex]P(0.35 <X < 0.55) =P( Z< 2.6595 ) - P( Z < -2.6596 )[/tex]

From the z-table  

    [tex]P( Z< 2.6595 ) = 0.99609[/tex]

and

    [tex]P( Z< - 2.6595 ) = 0.0039128[/tex]

So

   [tex]P(0.35 <X < 0.55) =0.99609 - 0.0039128[/tex]

=> [tex]P(0.35 <X < 0.55) = 0.9921772[/tex]

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