To find the 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, we can use Chebyshev's inequality.

Given:

• Population $P \sim \text{Uniform}[0, 12]$
• Sample size $n = 50$
• The tolerance level $c = 3$

Step-by-Step Solution

1. Calculate the Population Mean and Variance:

   • For a uniform distribution $U[a, b]$, the mean $\mu$ and variance $\sigma^2$ are given by: $\mu = \frac{a + b}{2}, \quad \sigma^2 = \frac{(b - a)^2}{12}$
   • In our case, $a = 0$ and $b = 12$: $\mu = \frac{0 + 12}{2} = 6$ $\sigma^2 = \frac{(12 - 0)^2}{12} = \frac{144}{12} = 12$

2. Calculate the Variance of the Sample Mean $\bar{X}$:

   • The variance of the sample mean $\bar{X}$ is $\sigma^2 / n$: $\text{Var}(\bar{X}) = \frac{12}{50} = 0.24$

3. Apply Chebyshev’s Inequality:

   • Chebyshev’s inequality states that for any $k > 0$, $P(|\bar{X} - \mu| \geq k) \leq \frac{\text{Var}(\bar{X})}{k^2}$
   • We want the probability that the sample mean $\bar{X}$ is within 3 of the actual mean, so we set $k = 3$: $P(|\bar{X} - \mu| < 3) = 1 - P(|\bar{X} - \mu| \geq 3) \geq 1 - \frac{0.24}{3^2}$ $= 1 - \frac{0.24}{9} = 1 - 0.0267 = 0.9733$

Answer:

The lower bound on the probability that the sample mean will be at most 3 away from the actual mean is approximately: $\boxed{0.973}$

Would you like further details or have any other questions?

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1. How would the probability change if the sample size increased?
2. What happens to the bound if the population variance increases?
3. How would this result change if we used a different inequality?
4. Can the weak law of large numbers provide an exact probability in any case?
5. What is the impact of changing the tolerance level $c$?

Tip: Chebyshev’s inequality is powerful for any distribution shape, providing a conservative probability bound.