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 = 
~ N(0,1)
where, 
= % of Indianapolis residents who were rooting for the Giants in Super Bowl XLVI in a sample of 200 residents = 
= 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( < 0.85)
P( < 0.85) = P( < 
) = P(Z < -1.98) = 1 - P(Z 
1.98)
= 1 - 0.97615 = 0.02385
<em>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.</em>
<em />
Therefore, probability that fewer than 170 were rooting for the Giants in Super Bowl XLVI is 0.02385.
