Approximate the probability that the sum of the 16 independent uniform (0,1) random variables exceeds 10.
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 of a large number of independent and identically distributed random variables will approximately follow a normal distribution.
In this case, since we have 16 independent uniform (0,1) random variables, the sum will follow a normal distribution with a mean of 16 * (mean of uniform distribution) = 16 * (0.5) = 8, and a standard deviation of sqrt(16 * (variance of uniform distribution)) = sqrt(16 * (1/12)) = sqrt(4/3) = 2/sqrt(3) ≈ 1.155.
To find the probability that the sum exceeds 10, we want to find P(S > 10), where S represents the sum of the 16 random variables.
First, we standardize the value of 10 using the formula z = (x - μ) / σ, where x is the value we want to standardize, μ is the mean, and σ is the standard deviation.
For 10, z = (10 - 8) / (2/sqrt(3)) ≈ 1.732.
Next, we use a standard normal table or a calculator to find the probability associated with this z-score.
Using a standard normal table, we can find the probability that the sum exceeds 10 by looking up the z-score 1.732. From the table, we can see that the area to the right of 1.73 is approximately 0.0418.
Therefore, the approximate probability that the sum of the 16 independent uniform (0,1) random variables exceeds 10 is approximately 0.0418 or 4.18%.
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 if we have a large enough sample size and the variables are independent, then the sum of those variables will be approximately normally distributed.
In this case, we have 16 independent uniform (0,1) random variables, which means that each variable has a probability density function that is a straight horizontal line between 0 and 1.
To solve this problem, we can calculate the mean and standard deviation of the sum of these 16 random variables.
The mean of a uniform (0,1) random variable is (0 + 1) / 2 = 0.5.
The variance of a uniform (0,1) random variable is (1 - 0)^2 / 12 = 1/12.
Since we have 16 independent random variables, the mean of the sum is 16 * 0.5 = 8, and the variance of the sum is 16 * 1/12 = 4/3.
To find the probability that the sum exceeds 10, we need to calculate the z-score and then find the corresponding probability using a standard normal distribution table.
The z-score is calculated as (x - mean) / standard deviation, where x is the value 10.
(z-score) = (10 - 8) / sqrt(4/3) = 2 / sqrt(4/3) ≈ 2.3094
Using a standard normal distribution table, we can find that the probability of a z-score greater than 2.3094 is approximately 0.0107.
Therefore, the approximate probability that the sum of the 16 independent uniform (0,1) random variables exceeds 10 is 0.0107, or about 1.07%.