Question

recall that chebyshevs theorem states that for any set of data and for any constant k greater than 1, the proportion of the data that must lie within k standard deviations on either side of the mean is at least 1 - \\(\frac{1}{k^{2}}\\). therefore, for data with mean \\(\mu\\) and standard deviation \\(\sigma\\), at least 1 - \\(\frac{1}{k^{2}}\\) of data must fall between \\(\mu - k\sigma\\) and \\(\mu + k\sigma\\). when k = 3, we have the following. 1 - \\(\frac{1}{k^{2}}\\) = 1 - \\(\frac{1}{3^{2}}\\) = 1 - \\(\frac{1}{9}\\) = \\(\frac{8}{9}\\) or 88.9%. therefore, for any set of data, at least 88.9% of the data must fall between \\(\mu - 3\sigma\\) and \\(\mu+(\space)\sigma\\), or, in other words, within ( ) standard deviations of the mean.

Explanation:

Step1: Substitute k value

We substitute $k = 3$ into the formula $1-\frac{1}{k^{2}}$. So we get $1-\frac{1}{3^{2}}=1 - \frac{1}{9}$.

Step2: Calculate the result

$1-\frac{1}{9}=\frac{9 - 1}{9}=\frac{8}{9}\approx0.889 = 88.9\%$.
For the last blank, since we are using $k = 3$ in the Chebyshev's formula $\mu\pm k\sigma$, when considering $\mu + k\sigma$, the value of $k$ is 3. And we are looking at the number of standard - deviations from the mean, which is 3.

Answer:

$1-\frac{1}{9}=\frac{8}{9}$; 3; 3