Respuesta :
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 = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = % of Indianapolis residents who were rooting for the Giants in Super Bowl XLVI in a sample of 200 residents = [tex]\frac{170}{200}[/tex] = 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([tex]\hat p[/tex] < 0.85)
P([tex]\hat p[/tex] < 0.85) = P( [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]\frac{0.85-0.90}{\sqrt{\frac{0.85(1-0.85)}{200} } }[/tex] ) = P(Z < -1.98) = 1 - P(Z [tex]\leq[/tex] 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.