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The period of a pendulum is measured 16 times. The average value of the period over these 16 trials is calculated to be 1.50 seconds, whereas the standard deviation is found to be 0.24 seconds. The standard error (or standard deviation of the mean) therefore is 0.24 seconds divided by the square root of 16, yielding 0.06 seconds. Assuming that the uncertainty of this period arises solely from random error for these 16 trials, about how many additional trials would need to be completed to reduce the standard error to 0.03 seconds? ______________

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onyebuchinnaji

Answer:

The additional trials needed is 48 trials

Explanation:

Given;

initial number of trials, n = 16 trials

the standard deviation, σ = 0.24 s

initial standard error, ε = 0.06 s

The standard error is given by;

[tex]\epsilon = \frac{\sigma}{\sqrt{n} }[/tex]

To reduce the standard error to 0.03 s, let the additional number of trials = x

[tex]0.03= \frac{0.24}{\sqrt{n+x} } \\\\0.03= \frac{0.24}{\sqrt{16+x} }\\\\0.03\sqrt{16+x} = 0.24\\\\\sqrt{16+x} = \frac{0.24}{0.03} \\\\\sqrt{16+x} = 8\\\\16+x = 8^2\\\\16+x = 64\\\\x = 64 -16\\\\x = 48 \ trials[/tex]

Therefore, the additional trials needed is 48 trials.