Question

A random variable X has a mean μ=10 and a variance σ

2
=4. Using Chebyshev's theorem, find (a) P(∣X−10∣≥3) (b) P(∣X−10∣<3) (c) P(5

Answer 1

The probability that a random variable X is at least 3 units away from its mean 10 is at most 4/9, using Chebyshev's theorem. The probability of X being less than 3 units away from its mean is therefore 5/9. The third part of the question is incomplete and cannot be answered.

Step-by-step explanation:

The student's question involves using Chebyshev's theorem to find probabilities related to a random variable with a given mean and variance. We are given a mean (μ) of 10 and a variance (σ2) of 4. Chebyshev's theorem is applicable to any probability distribution and provides bounds on the probability that a random variable is a certain number of standard deviations away from the mean. Using this theorem, we can calculate the following:

1. P(|X-10|≥3): Chebyshev's theorem states that for any k>1, P(|X-μ|≥kσ) ≤ 1/k2. In this case, k=3/σ where σ is the standard deviation. Since the variance is 4, σ is 2. So, k=3/2 and P(|X-10|≥3) ≤ 1/(3/2)2 = 4/9.
2. P(|X-10|<3): This probability is complementary to the probability found in part (a), so P(|X-10|<3) = 1 - P(|X-10|≥3). Substituting the computed value from part (a), we get P(|X-10|<3) = 1 - 4/9 = 5/9.
3. The third part of the question is incomplete and therefore cannot be answered as provided.