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Math Help - Probability / Statistics

  1. #1
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    Probability / Statistics

    I was wondering if someone could check my work thanks so the question is like so:
    The time it takes for a pendulum to complete the one period is simplified by the formula: T=2 pi square root (L/g) Where g is the constant measuring 9.80655m/s^2 and L is the length of the pendulum. The length of the pendulums from a manufacture are normally distributed with a mean of 10cm and a standard deviation of 0.01 cm. Assuming the at a 10cm pendulum gives the correct time, what is the probability that a clock using one of theses pendulums will lose more than 1 minute a day?

    so my work so far:



    Standard deviation: 0.01cm
    Mean= 10cm
    T= 2pi [sqrt] L/g
    Time increases by: 1/ 24x 60

    P(x> 10.014)
    = p (z> (10.014-10 / 0.01)
    = P (Z > 1.4)
    Z= 0.919243


    my question is this correct? or do I minus one from thz?

    thanks!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by frogsrcool View Post
    I was wondering if someone could check my work thanks so the question is like so:
    The time it takes for a pendulum to complete the one period is simplified by the formula: T=2 pi square root (L/g) Where g is the constant measuring 9.80655m/s^2 and L is the length of the pendulum. The length of the pendulums from a manufacture are normally distributed with a mean of 10cm and a standard deviation of 0.01 cm. Assuming the at a 10cm pendulum gives the correct time, what is the probability that a clock using one of theses pendulums will lose more than 1 minute a day?

    so my work so far:



    Standard deviation: 0.01cm
    Mean= 10cm
    T= 2pi [sqrt] L/g
    Time increases by: 1/ 24x 60
    ……
    P(x> 10.014)
    = p (z> (10.014-10 / 0.01)
    = P (Z > 1.4)
    Z= 0.919243


    my question is this correct? or do I minus one from thz?

    thanks!
    If the clock loses more than 1 min /day the period of the pendulum is too
    long by a factor of 1+1/(24 \times 60) \approx 1.000694

    The period of the nominal pendulum is:

    <br />
T_{nom} = 2\pi \sqrt{\frac{L}{g}}=2\pi \sqrt{\frac{0.1}{9.80655}}\approx 0.634486 \mbox{ s}<br />

    The period of the pendulum in a clock that loses more than 1 min/day is
    greater than:

    <br />
1.000694 \times T_{nom}=0.634926 \mbox{ s}<br />
.

    So the pendulum is longer than:

    <br />
L=\left[ \frac{0.634926}{2 \pi} \right]^2 \times 9.80655 \approx 0.100139 \mbox{ m}<br />

    Now the SD of the length of pendulums is 0.01 \mbox{ cm} or 0.0001 \mbox{ m}. So now we compute the z-score of L:

    <br />
z=(0.100139-0.1)/0.0001 = 1.39<br />

    which we look up in our table of the cumulative standard normal to find that
    the probability that this is not exceeded is 0.9177, so the probability that
    t is exceeded is 1-0.9177 = 0.0823, which is the probability that the clock
    loses more than 1 minute per day.

    RonL
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