Question

What is the probability that from a sample of 100 Indianapolis residents, fewer than 95% were rooting for the Giants in Super Bowl XLVI

Answer 1

Complete Question

Super Bowl XLVI was played between the New York Giants and the New England Patriots in Indianapolis. Due to a decade-long rivalry between the Patriots and the city’s own team, the Colts, most Indianapolis residents were rooting heartily for the Giants. Suppose that 90% of Indianapolis residents wanted the Giants to beat the Patriots. What is the probability that from a sample of 100 Indianapolis residents, fewer than 95% were rooting for the Giants in Super Bowl XLVI?

Answer:

The value is  $P(X < 0.95) = 0.95224$

Step-by-step explanation:

From the question we are told that

  The proportion that wants Giants to win  p = 0.90  

   The sample size is n  =  100

 Generally the mean of this sampling distribution is  

       $\mu_{x} = 0.90$

Generally the standard deviation is  

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

=>   $\sigma = \sqrt{\frac{ 0.90 (1- 0.90)}{100} }$

=>   $\sigma = 0.03$

Generally the probability that  from a sample of 100 Indianapolis residents, fewer than 95% were rooting for the Giants in Super Bowl XLVI is mathematically represented  as

     $P(X < 0.95) = P(\frac{ X - \mu_{x}}{\sigma} < \frac{0.95 -0.90}{0.03} )$

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

       $P(X < 0.95) = P(Z < 1.667 )$

=>    $P(X < 0.95) = 0.95224$