Question

The mean age of participation is 46 with a standard deviation of 3.2. The minimum percentages of age participation between 41.2 and 50.8 is ______________. Apply Chebyshev's Rule and round to the nearest hundredth of a percent.Fill in the blank.

Answer 1

Using Chebyshev's rule, we can write

$P(|X-\mu| \leq k\sigma ) \geq 1 - \frac{1}{k^2}$

Now,

$P(41.2 \leq X \leq 50.8) = P(|X-46|\leq 1.5*3.2) = P(|X-\mu|\leq 1.5\sigma )$

$\Rightarrow P(41.2 \leq X \leq 50.8) = P(|X-\mu|\leq 1.5\sigma ) \geq 1 - \frac{1}{1.5^2}$

$\Rightarrow P(41.2 \leq X \leq 50.8) \geq 0.555556$ (ans)