Suppose a random variable X has mean μX = 6.3, and variance V(X) = 2.2. According to Chebyshev's Theorem, what is the upper bound for the probability that the distance between the value of X and μ will be at least 4.8?
I don't even know where to start. But If I can see the steps to complete this problem, it would help out tremendously with the others.
To use Chebyshev's Theorem, we need to find the value of k, which represents the number of standard deviations away from the mean. The theorem states that for any value k greater than 1, the probability that a randomly selected data point falls within k standard deviations from the mean is at least 1 - 1/k^2.
In this case, we want to find the upper bound for the probability that the distance between the value of X and μ will be at least 4.8. We can set up the equation as follows:
P(|X - μ| ≥ 4.8) ≤ 1/k^2
We know that the variance (V(X)) is equal to the square of the standard deviation (SD(X)). Therefore, we can find SD(X) by taking the square root of the variance:
SD(X) = sqrt(2.2)
Next, we can substitute the value of k in terms of standard deviation:
k = (4.8 / SD(X))
Now we can substitute the value of k into the equation:
P(|X - μ| ≥ 4.8) ≤ 1 / [(4.8 / SD(X))^2]
P(|X - μ| ≥ 4.8) ≤ 1 / [(4.8^2) / (SD(X))^2]
P(|X - μ| ≥ 4.8) ≤ (SD(X))^2 / (4.8^2)
P(|X - μ| ≥ 4.8) ≤ (2.2) / (4.8^2)
P(|X - μ| ≥ 4.8) ≤ 2.2 / 23.04
P(|X - μ| ≥ 4.8) ≤ 0.095
Therefore, the upper bound for the probability that the distance between the value of X and μ will be at least 4.8 is 0.095, or 9.5%.