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Thread: Please help .. z scores ...

  1. #1
    Dec 2009

    Question Please help .. z scores ...

    1) Processing of the ceramic alloy involves a sintering ( firing ) process which produces a dense hard material. the sintering operation had to be carefully monitored and the tolerance set for the sentering temperature is 1200 c + - 10 c with a standard deviation of 3.6 c. For the investigative work the temperature must be that the probability of the temperature drifting outside tolerance must not exceed 1%.

    Using the temperature data supplied assess the temperature capability for the sintering process of trial material.

    For this problem i have tried to do the Z score but i do not understand where the mean is .. I think i got the Z which < 0.01 - 1% please help me Do i need to calculate the mean ? please help

    Z = X-μ / σ μ- Mean σ Standard deviation

    2. In order to launch the new product , special samples kits will be produces . The cost of production will be high initially so it is important not to waste material. However , it is important that the manchines which dispense the modelling and liner liquids do not underfill the containers .
    The modelling liquid bottle states a nominal volume of 100 ml . Measeruments you have taken on modelling the liquid volume going into the bottles indicate a mean of 100 and standerd deviation of 0.2 ml. What amount of liquid should be dispensed to limit the risk of underfilling to 1 in 40 ?
    For this problem i have calculated The X scores as i got the Z score , than 1/40
    0.065 = 0.025 I have checked it on the table of normal destribuition and it falls on between 0.06 and 0.07

    1/40 - 0.025 - 2.5 %
    P(Z<-0.065)=0.025 P(Z< (X-μ)/σ = 0.025

    -0.065= X-100/0.2
    -0.065= X/0.2 - 500
    -.0065 +500 = x/0.2
    499.935 = X/0.2
    X =499.935 times 0.2
    x =99.987

    So for the ascwer of the question 2 its should be dispensed 99.987 ml of the 2.5 % to limit the risk of underfilling

    Is the answer correct for this problem .. Please help
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  2. #2
    Super Member
    Jul 2009
    Read, read again, and then re-read the problem if it is giving you a headache.

    A. They give you all the information you need. You are told that a process (who cares what the process is, basically just skip the entire first sentence of this question - they are trying to overload you with information to see if you can actual comprehend what is being asked on a fundemental level) is fine if it produces results that are within 10C's of 1200. Thus random variables ranging from 1190 to 1210 will be considered successful. The question that want to know with what percentage will the process produce good results. Well obviously it will produce good results if the values of your samples run between 1190 and 1210. You then, using Z-scores (I assume the distribution of temperatures are normal since the process technically has an infinite population), find the area bounded between 1190 and 1210.

    B. Is fine.
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