# Probability / Statistics

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• Jun 10th 2007, 12:22 PM
frogsrcool
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!
• Jun 10th 2007, 02:27 PM
CaptainBlack
Quote:

Originally Posted by frogsrcool
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:

$
T_{nom} = 2\pi \sqrt{\frac{L}{g}}=2\pi \sqrt{\frac{0.1}{9.80655}}\approx 0.634486 \mbox{ s}
$

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

$
1.000694 \times T_{nom}=0.634926 \mbox{ s}
$
.

So the pendulum is longer than:

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

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 $L$:

$
z=(0.100139-0.1)/0.0001 = 1.39
$

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