Question 673803: The 2010 American Census reported that 28.6% of American teens held part time jobs. Let p = .286 represent the proportion of American teens that have jobs. Find, if possible, the probability that a random sample of 30 American teens will have fewer than 20% who have part time jobs.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The 2010 American Census reported that 28.6% of American teens held part time jobs. Let p = .286 represent the proportion of American teens that have jobs. Find, if possible, the probability that a random sample of 30 American teens will have fewer than 20% who have part time jobs.
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z(0.20) = (0.20-0.286)/sqrt[0.286*0.714/30] = -1.0424
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P(p-hat < 0.20) = P(z < -1.0424) = normalcdf(-100,-1.0424) = 0.1486
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Cheers,
Stan H.
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