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December 22, 2014

December 22, 2014

Posted by **z score!** on Sunday, June 10, 2007 at 3:20pm.

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!

For P(Z > 1.4), take 1 minus your look-up table value of .9192. So P()= .0808

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