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

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!

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

## Answer this Question

## Related Questions

- physics - The period of a simple pendulum is measured to be 4.0 seconds in a ...
- physics - The period of a simple pendulum is measured to be 4.0 seconds in a ...
- physics - The period of a simple pendulum is measured to be 4.0 seconds in a ...
- Math - Hi! So this question kind of has two parts to it. If someone could help ...
- math - The formula T=2*pi*sqrt[L/g] gives the period of a pendulum of length l ...
- College Algebra - The period (the time required for one complete swing) of a ...
- math - The length of time that it takes for a pendulum to make one complete ...
- algebra - The Length of time (T) in seconds it takes the pendulum of a clock to ...
- Math - The length of time (T) in seconds it takes the pendulum of a clock to ...
- math - The formula T=2*pi*sqrt[L/32] gives the period of a pendulum of length l...

More Related Questions