cooldudeman
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Stuck on 3a. Im not too sure on what a point estimate is. I thought it would just be 10.01ounces given in the question.

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Smaug123
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(Original post by cooldudeman)
Stuck on 3a. Im not too sure on what a point estimate is. I thought it would just be 10.01ounces given in the question.
I think you're right, but the question is a bit odd. What answer do you have for the probability that it's the true mean? (That answer is why I think the question is odd.)
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cooldudeman
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(Original post by Smaug123)
I think you're right, but the question is a bit odd. What answer do you have for the probability that it's the true mean? (That answer is why I think the question is odd.)
Im not really sure. Im thinking that it would be the reverse of finding the confidence interval. Id guess its 50% because of the mean being in the middle of the normal dist curve.

So on normal tables for P(Z<0)
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Smaug123
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(Original post by cooldudeman)
Im not really sure. Im thinking that it would be the reverse of finding the confidence interval. Id guess its 50% because of the mean being in the middle of the normal dist curve.

So on normal tables for P(Z<0)
What is Probability(mu = 10.01) where mu follows a normal distribution? (It does, since the sum of normally distributed random variables is normally distributed.)
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sophieshoesmith
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OUNCES! How many years old is that question!
I guess its asking whats probabilitity that 10.005>u>10.015 and thats a width of .01/sqr(0.0004) (ie 0.01/0.006 ie 1.5 standard deviations)

so 0.75 s.ds. each side of the mean does cover about 50%
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cooldudeman
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(Original post by Smaug123)
What is Probability(mu = 10.01) where mu follows a normal distribution? (It does, since the sum of normally distributed random variables is normally distributed.)
So 'point' means its equal to it im guessing

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Smaug123
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(Original post by cooldudeman)
So 'point' means its equal to it im guessing

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P(Z = 0) = 0. The probability that a continuous random variable takes any given value is 0.
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cooldudeman
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(Original post by Smaug123)
P(Z = 0) = 0. The probability that a continuous random variable takes any given value is 0.
I thought continuity corrections apply?

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Smaug123
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(Original post by cooldudeman)
I thought continuity corrections apply?

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Continuity corrections apply to discrete distributions being approximated by continuous ones. We are only considering normal distributions here.
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cooldudeman
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(Original post by Smaug123)
Continuity corrections apply to discrete distributions being approximated by continuous ones. We are only considering normal distributions here.
So the probability is 0%? Thats so strange

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Smaug123
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(Original post by cooldudeman)
So the probability is 0%? Thats so strange

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It's because of the fact that \int_a^a f(x) dx = 0.
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cooldudeman
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(Original post by Smaug123)
It's because of the fact that \int_a^a f(x) dx = 0.
i was kinda saying strange because it makes the other questions on 3 strange to do. for 3b, to find the standard error, you do the Z value multiplied by the standard deviation but that would be zero...
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