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 200 Indianapolis residents, fewer than 170 were rooting for the Giants in Super Bowl XLIV?

Answer 1

Answer:

Probability that from a sample of 200 Indianapolis residents, fewer than 170 were rooting for the Giants in Super Bowl XLVI is 0.02385.

Step-by-step explanation:

We are given that 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.

Let p = % of Indianapolis residents wanted the Giants to beat the Patriots = 90%

The z-score probability distribution for proportion is given by;

                   Z = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where,  $\hat p$ = % of Indianapolis residents who were rooting for the Giants in Super Bowl XLVI in a sample of 200 residents = $\frac{170}{200}$ = 0.85

           n = sample of residents = 200

So, probability that from a sample of 200 Indianapolis residents, fewer than 170 were rooting for the Giants in Super Bowl XLVI is given by = P($\hat p$ < 0.85)

     P($\hat p$ < 0.85) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\frac{0.85-0.90}{\sqrt{\frac{0.85(1-0.85)}{200} } }$ ) = P(Z < -1.98) = 1 - P(Z $\leq$ 1.98)

                                                                   = 1 - 0.97615 = 0.02385

The above probability is calculated using z table by looking at value of x = 01.98 in the z table which have an area of 0.97615.

Therefore, probability that fewer than 170 were rooting for the Giants in Super Bowl XLVI is 0.02385.