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it has been estimated that 85% of all people in San Diego watched the Chargers in...

it has been estimated that 85% of all people in San Diego watched the Chargers in the Super Bowl. If 300 randomly selected people in San Diego are asked if they watched the chargers, what is the probability that more than 90% say yes?

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Answer #1

Using central limit theorem to proportions,

P( < p ) = P( Z < - p / sqrt( p ( 1 - p) / n) )

So,

P( > 0.90) = P( Z > 0.90 - 0.85 / sqrt( 0.85 * 0.15 / 300) )

= P (Z > 2.4254)  

= 0.0076 (Probability calculated from Z table)

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