Chebyshev's inequality states that for any distribution, the proportion of outcomes that lie within $k$ standard deviations from the mean is at least $1 - \frac{1}{k^2}$. Specifically, it tells us the minimum probability that a random variable will fall within a certain range around the mean.

The formula is: $P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}$ where:

• $\mu$ is the mean,
• $\sigma$ is the standard deviation,
• $k$ is the number of standard deviations from the mean.

Step-by-step solution:

Given:

• Mean $\mu = 60$,
• Standard deviation $\sigma = 8$,
• The range is from $-8$ to $128$.

Step 1: Find the number of standard deviations, $k$, that corresponds to the boundaries.

We calculate how far the boundaries $-8$ and $128$ are from the mean in terms of standard deviations:

• For $-8$: $k_1 = \frac{| -8 - 60 |}{8} = \frac{68}{8} = 8.5$
• For $128$: $k_2 = \frac{| 128 - 60 |}{8} = \frac{68}{8} = 8.5$ Thus, $k = 8.5$.

Step 2: Apply Chebyshev's inequality.

Now that $k = 8.5$, Chebyshev's inequality gives the minimum probability that a random variable lies within $k$ standard deviations of the mean as: $P(|X - \mu| < k\sigma) \geq 1 - \frac{1}{k^2}$ Substituting $k = 8.5$: $P(|X - 60| < 8.5 \times 8) \geq 1 - \frac{1}{(8.5)^2} = 1 - \frac{1}{72.25}$ $P(|X - 60| < 8.5 \times 8) \geq 1 - 0.0138 = 0.9862$

Final Answer:

The minimum probability that the outcome is between $-8$ and $128$ is approximately $0.9862$.

Would you like more details on how Chebyshev's inequality works or have any other questions?

Here are 5 related questions for further exploration:

1. How does Chebyshev’s inequality compare with the Empirical Rule for normal distributions?
2. Can Chebyshev’s inequality be used for distributions that are not symmetric?
3. What is the probability for a smaller range, say between 40 and 80?
4. How would this problem change if the standard deviation were doubled?
5. How can Chebyshev’s inequality be applied in real-world data analysis?

Tip: Chebyshev's inequality is useful when you don't know the exact shape of a distribution—it works for any distribution!