First find the distribution of sample mean Xbar
Xbar has a normal distribution with mean = 50, std = 5/sqrt(100)=0.5
P(Xbar>51) = P(Z > (5150)/0.5) = P(Z>2) = 1P(Z<2) = 1 0.9772 = 0.0228
5. Did you need to use the Central Limit theorem to answer #4? Why or why not?
No, because the population is normal. For a normal population,
the sampling
distribution of Xbar is normal for all values of n.
The annual rainfall in Cleveland, Ohio has an unknown distribution with mean 40.2
inches and standard deviation 8.4 inches
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What is the probability that next year's rainfall will be less than 40 inches?
Can’t get the probability from the problem information because the population
distribution is unknown here.
7.
What is the probability that the average yearly rainfall over 30 years (selected at
random) will be less than 40 inches?
By CLT, since n=30 big enough, the sampling distribution of Xbar would be
approximately normal with mean = 40.2 and std= 8.4/sqrt(30)= 1.53
P(Xbar<40 )= P(Z<(4040.2)/1.53) = P(Z < 0.13) = 0.4483
8. Did you need to use the Central Limit Theorem for #7?
Why or why not?
Yes. Because the population distribution is unknown here, so we need CLT to get the
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 Spring '07
 rumseyjohnson
 Statistics, Central Limit Theorem, Normal Distribution, Probability, Standard Deviation

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