A random sample of size 50 is collected from a population 𝑃P, where 𝑃∼P∼ Uniform[0,12]. Find a lower bound on the probability that the sample mean will be at most 3 away from the actual mean using the weak law of large numbers. Enter the answer correct to three decimal places.
Question
A random sample of size 50 is collected from a population 𝑃P, where 𝑃∼P∼ Uniform[0,12]. Find a lower bound on the probability that the sample mean will be at most 3 away from the actual mean using the weak law of large numbers. Enter the answer correct to three decimal places.
Solution
The weak law of large numbers states that for a random sample of size n from a population with a finite mean μ and finite variance σ^2, the sample mean will converge in probability to the population mean as n approaches infinity.
In this case, we have a uniform distribution on the interval [0,12], so the population mean μ is (0+12)/2 = 6. The population variance σ^2 is ((12-0)^2)/12 = 12.
The standard deviation of the sample mean, which is the standard deviation of the population divided by the square root of the sample size, is sqrt(12/50) = 0.49 approximately.
We want to find a lower bound on the probability that the sample mean is at most 3 away from the actual mean. This is equivalent to finding the probability that the absolute difference between the sample mean and the population mean is less than or equal to 3.
By Chebyshev's inequality, this probability is at least 1 - (σ^2/(n*k^2)), where k is the number of standard deviations away from the mean. In this case, k = 3/0.49 = 6.12 approximately.
So the probability is at least 1 - (12/(50*6.12^2)) = 1 - 0.032 = 0.968.
Therefore, the lower bound on the probability that the sample mean will be at most 3 away from the actual mean is 0.968, or 0.968 when rounded to three decimal places.
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