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Math Help - Sampling Distributions

  1. #1
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    Sampling Distributions

    The thickness (mm) of the coating applied to disk drives is a characteristic that determines the usefulness of the product. When no unusual circumstances are present, the thickness
    X has a normal distribution with mean of 3 mm and standard deviation of 0.05 mm. Suppose that the coating process will be monitored by selecting a random sample of 16 drives from each shift's production and determining the sample mean coating.
    (a) Describe the sampling distribution of the sample mean.


    (b) When no unusual circumstances are present, we expect the sample mean
    to be within 3SE (standard error) of 3 mm, the desired coating thickness value. A value of the sample meanfarther from 3 than 3SEis interpreted as an indication of a problem that needs attention. Compute 3+/-3SE.

    (c) Referring to part (b), what is the probability that the sample meanwill be outside 3+/-3SEjust by chance (that is, when there are no unusual circumstances)?

    (d) Suppose that a machine used to apply the coating is out of adjustment, resulting in a mean coating thickness of 3.05 mm. What is the probability that a problem will be detected when the next sample is taken?
    (Hint: This will occur if sample mean > 3+3SEor sample mean < 3-3SEwhen mu = 3.05).

    So for (a) I wrote the sample mean is 3mm, and the SE of the sample mean is 0.0125.

    For (b) 3+/-3(SE) = 2.9625, 3.0375

    For (c) P(sample mean > 3+3SEor sample mean < 3-3SE) = 0.0026

    For (d) I'm thinking that it's the same calculation as in (c) except I should use 3.05 when finding the z-score. For example,

    z = (3.0375-3.05)/0.0125 = -1
    z = (2.9625 - 3.05)/0.0125 = -7

    Then it becomes P(z < -7) + [1 - P(z < -1)]

    Is this right? If so, what do I write for P(z < -7), since it would be a very minuscule number???

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  2. #2
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    Rounding at 4DP

    I concur with what you have.

    From there you hopefully find

    P(z > -1)
    = (1 - P(z < -1)) to be
    = 1 - (1 - P(z < 1))
    = P(z < 1)
    = 0.8413 (4dp)

    You are correct that the other term will be minuscule.
    in fact P(z < 4) = 1.0000 (4dp) from my tables
    so P(z < -4) = 0.0000 (4dp)

    Thus we can be very secure that to 4 decimal places P(z < -7)
    will contribute a negligible amount

    So we can write
    P(z < -7) + [1 - P(z < -1)] = 0.0000 + 0.8413 = 0.8413
    with a degree of confidence

    However, if you wanted to make the point that it may just tip over the value to 0.8414 to 4DP (which seems very unlikely in this case)
    you could always write with absolute certainty that:
    P(z < -7) + [1 - P(z < -1)] = 0.841
    to 3DP
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  3. #3
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    Re: Sampling Distributions

    Hey can you please explain your answers in depth! I'm stuck on this question! Thanks!
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