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    The lengths of bolts produced by a machine have an unknown distribution with mean 3.03 cm and standard deviation 0.20cm. A sample of 100 bolts is taken.

    What size sample is required if the probability that the mean is less than 3cm is to be less than 1%?

    Can anyone show me the working?
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    By the CLT the distribution of the sample mean Xbar is approx normal with mean=3.03 and stddev= (0.2^2)/n.

    So Prob( Xbar < 3 ) <0.01 is what you need to solve.
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    (Original post by vc94)
    By the CLT the distribution of the sample mean Xbar is approx normal with mean=3.03 and stddev= (0.2^2)/n.

    So Prob( Xbar < 3 ) <0.01 is what you need to solve.
    I solved the problem in the same way. But my answer is that N>239.2 while the solution says it is N>240. So I wonder what's the problem of my calculation.
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    Just use a Normal distribution. Since it qualifies for Central Limit theorem (n>29). Simply work backwards and find the Z value for 1%.
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    (Original post by jimmy_law)
    I solved the problem in the same way. But my answer is that N>239.2 while the solution says it is N>240. So I wonder what's the problem of my calculation.
    Prob ( Z< (3-3.03)/(0.2/sqrt(N)) )<0.01

    (3-3.03)/(0.2/sqrt(N)) < -2.3263

    ... N> 240.5187

    So N>240
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    (Original post by vc94)
    Prob ( Z< (3-3.03)/(0.2/sqrt(N)) )<0.01

    (3-3.03)/(0.2/sqrt(N)) < -2.3263

    ... N> 240.5187

    So N>240
    Oh, I simply used -2.32 instead of -2.3263.
    It seems that I have to be more precise.
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    Sorry but one more question.

    A fair die is rolled 35 times. Find the approximate probability that the total of the 35 scores is less than 100.

    I have totally no idea when I tackle this question.
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    (Original post by jimmy_law)
    Sorry but one more question.

    A fair die is rolled 35 times. Find the approximate probability that the total of the 35 scores is less than 100.

    I have totally no idea when I tackle this question.
    I'd imagine you'd set up an approx. normal distribution of X being one dice throw in a sample of 35 then divide a 100 by 35 and find the chance that it is less than that.

    Or set up another normal with 35 rolls instead of one then you won't have to divide by 35.
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    (Original post by StephenP91)
    I'd imagine you'd set up an approx. normal distribution of X being one dice throw in a sample of 35 then divide a 100 by 35 and find the chance that it is less than that.

    Or set up another normal with 35 rolls instead of one then you won't have to divide by 35.
    So the mean and variance of the normal approximations will be 35 times of the E(X) and Var(X)?
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    (Original post by jimmy_law)
    So the mean and variance of the normal approximations will be 35 times of the E(X) and Var(X)?
    Yeah.
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    (Original post by StephenP91)
    Yeah.
    Oops, the variance should be 35^2 (Var(X))...sorry

    but I still can't get the correct answer, which is 0.013..
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    (Original post by jimmy_law)
    Oops, the variance should be 35^2 (Var(X))...sorry

    but I still can't get the correct answer, which is 0.013..
    It wouldn't be squared.

    Use my first method, you get the right answer.
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    (Original post by StephenP91)
    It wouldn't be squared.
    why?
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    (Original post by jimmy_law)
    why?
    Do this:

    Ex of all dice throws (6 sided fair die) is 3.5 and Varx is always 2.92. Set up a normal with Ex 3.5 and Varx 2.92/35.

    Then find P(x<100/35) and you'll get about 0.013.

    As for why you can't square you should have done in S2. You set up 35 normals and add them seperately other wise it'll be 35 of the same roll rather than 35 seperate rolls.

    If you can't remember the Varx or Ex you can set up a frequency table and calculate Ex and Ex squared and use the Varx formula.
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    T=X1+X2+...+X35
    Each Xi has a discrete uniform distribution with E(Xi)=3.5 and Var(Xi)=(5x7)/12 = 35/12

    So E(T)=122.5, Var(T)=(35^2)/12
    By the CLT, Tbar is approx normal with mean=3.5 and var=35/12, so T is normal.

    You want prob(T<100) ...use continuity correction since T has integer values...P(T<=99.5)
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    (Original post by StephenP91)
    Do this:

    Ex of all dice throws (6 sided) is 3.5 and Varx is always 2.92. Set up a normal with Ex 3.5 and Varx 2.92/35.

    Then find P(x<100/35) and you'll get about 0.013.

    As for why you can't square you should have done in S2. You set up 35 normals and add them seperately other wise it'll be 35 of the same roll rather than 35 seperate rolls.

    If you can't remember the Varx or Ex you can set up a frequency table and calculate Ex and Ex squared and use the Varx formula.
    Thank you very much!
 
 
 
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