SOLUTION: 1. At a house, there are 6 seniors, 4 juniors, and 2 sophomores. If a committee of 3 is selected at random, what is the probability that 1 sophomore and 2 seniors will be selected?

Question 205386: 1. At a house, there are 6 seniors, 4 juniors, and 2 sophomores. If a committee of 3 is selected at random, what is the probability that 1 sophomore and 2 seniors will be selected?
2. The average repair cost of a microwave oven is $55, with a standard deviation of $8. The costs are normally distributed. If 12 ovens are repaird, find the probability that the mean of the repair bills will be less than $51.
3. A researcher at Wilkes University wishes to estimate the average amount of money earned by all students who work summer jobs. The standard deviation is known to be $500. In a random sample of 60 Wilkes students who had summre jobs, the researcher found the mean amount of money they earned to be $3000. Find the 95% confidence interval for the mean amount of money earned by all students with summer jobs.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1. At a house, there are 6 seniors, 4 juniors, and 2 sophomores. If a committee of 3 is selected at random, what is the probability that 1 sophomore and 2 seniors will be selected?
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# of ways to get the committee as described: 2 = 2*15 = 30
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# of ways to pick the committee without condition: 12C3 = (12*11*10/(1*2*3)
= 220
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Probability of picking the committee as described = 30/220 = 3/22
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2. The average repair cost of a microwave oven is $55, with a standard deviation of $8. The costs are normally distributed. If 12 ovens are repaird, find the probability that the mean of the repair bills will be less than $51.
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z(51) = (51-55)/[8/sqrt(12)] = -1.732...
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P(x < 51) = P(z<-1.732) = 0.0416
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3. A researcher at Wilkes University wishes to estimate the average amount of money earned by all students who work summer jobs. The standard deviation is known to be $500. In a random sample of 60 Wilkes students who had summer jobs, the researcher found the mean amount of money they earned to be $3000. Find the 95% confidence interval for the mean amount of money earned by all students with summer jobs.
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sample mean = 3000
E = 1.96*500/sqrt(60)= 126.52
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95% CI : 3000-126.52 < u < 3000+126.52
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Cheers,
Stan H.

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