# I was wondering if someone could check my work thanks so the question is like so: The time it takes

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