I was wondering if someone could check my work thanks so the question is like so:

Question

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 1

Your calculations are almost correct. However, there is a minor mistake in calculating the value of Z. To find the probability that a clock using one of these pendulums will lose more than 1 minute a day, you need to calculate the Z-score correctly.

The Z-score represents how many standard deviations an observation is from the mean. In this case, you want to find the Z-score for a pendulum length of 10.014 cm.

Z = (x - μ) / σ

Where:
x = 10.014 cm (observation)
μ = 10 cm (mean)
σ = 0.01 cm (standard deviation)

Plug in these values to calculate the Z-score:

Z = (10.014 - 10) / 0.01
Z = 1.4

Now, to find the probability P(Z > 1.4), you need to take 1 minus the cumulative probability associated with the Z-score. In other words, find the area under the normal distribution curve to the right of the Z-score.

P(Z > 1.4) = 1 - P(Z ≤ 1.4)

You can look up the cumulative probability value for 1.4 in the Z-score table or use statistical software to find the value. Let's assume it is 0.9192, as you mentioned.

P(Z > 1.4) = 1 - 0.9192
P(Z > 1.4) = 0.0808

So the probability that a clock using one of these pendulums will lose more than 1 minute a day is approximately 0.0808, or 8.08%.