Probability / Statistics
posted by z score! on .
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
T= 2pi [sqrt] L/g
Time increases by: 1/ 24x 60
= p (z> (10.014-10 / 0.01)
= P (Z > 1.4)
my question is this correct? or do I minus one from thz?
For P(Z > 1.4), take 1 minus your look-up table value of .9192. So P()= .0808