Question

According to Chebyshev’s Theorem, at least what percent of the data values in a large data lie within 2.25 standard deviation to either side of the mean?

Answer 1

For any data set, the proportion (or percentage) of values that fall within  k  standard deviations from mean.

The percentage for 2.25

1-1/k^2

Interval (x-ks, x+ks)

Percentage = 1-1/(2.25)^2

= 80.24 %

Answer 2

Solution Using Chebyshev’s Theorem

Given:

• Number of standard deviations ($k$): 2.25

Chebyshev’s Formula:

$$\text{Minimum percentage within }k\text{ standard deviations}=1-\frac{1}{k^2}$$

Calculation:

$$1-\frac{1}{(2.25{)}^2}=1-\frac{1}{5.0625}\approx 1-0.1975=0.8025$$

Convert to Percentage:

$$0.8025\times 100=\boxed{{\displaystyle 80.25\mathrm{\%}}}$$

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Key Notes:

1. Chebyshev’s Theorem applies to any distribution (non-normal, skewed, etc.).

2. Contrast with Empirical Rule:

• For normal distributions, ~95% of data lies within ±2σ.

• Chebyshev’s gives a weaker but universal bound (here, ≥80.25% for $k=2.25$).