Approximate the probability that the sum of the 16 independent uniform (0,1) random variables exceeds 10.

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To approximate the probability that the sum of 16 independent uniform (0,1) random variables exceeds 10, we can use the central limit theorem.

The central limit theorem states that the sum (or average) of a large number of independent and identically distributed random variables with finite mean and standard deviation will be approximately normally distributed.

In this case, we have 16 independent uniform (0,1) random variables, which have a mean of 0.5 and a standard deviation of 1/√12. So the mean of the sum of these random variables would be 16 * 0.5 = 8, and the standard deviation would be 16 * (1/√12) = 4/√3 ≈ 2.31.

Now, we want to find the probability that the sum exceeds 10. We can standardize the value 10 using the mean and standard deviation we calculated:

Z = (X - μ) / σ
Z = (10 - 8) / 2.31 ≈ 0.87

Next, we can look up the probability associated with the Z-score of 0.87 in the standard normal distribution table. The table gives us the cumulative probability, so we need to find the complement of this probability to get the probability that the sum exceeds 10.

1 - P(Z < 0.87) ≈ 1 - 0.8078 = 0.1922

Therefore, the probability that the sum of the 16 independent uniform (0,1) random variables exceeds 10 is approximately 0.1922 or 19.22%.