Question

heights of adults women have a mean of 63.6 in and a standard deviation of 2.5 indies chebyshevs theorem say about the percentage of women with heights between 58.6 and 68.6in

Answer 1

Chebyshev's Theorem (Tchebysheff's Theorem) suggests that this percentage would be greater than
`75\%`.

Explanation:

Let
`Y` be a random variable with a mean of
`\mu` and a (finite) variance of
`\sigma`. By Chebyshev's Theorem, for any constant
`k` where
`k > 0`:

`\displaystyle P\left(\left|Y -\mu\right| \le k\, \sigma\right) \ge 1 - (1)/(k^2)`.

In this question, let
`Y` denote height (in inches.) Accordingly,
`\mu = 63.6` whereas
`\sigma = 2.5`.

Note, that the interval of interest
`(58.6,\, 68.6)` is centered at
`\mu = 63.6`. Besides, the difference between
`\mu` and either endpoints of this interval is equal to
`5.0`, which is the same as
`2\, \sigma`.

Rewrite the interval
`(58.6,\, 68.6)` as
`(63.6 - 2* 2.5,\, 63.6 + 2 * 2.5)`.

For the height
`Y` of one such individual,
`Y \in (58.6,\, 68.6)` is equivalent to
`\left|Y - 63.6\right| < 2 * 2.5`. With
`\mu = 63.6` whereas
`\sigma = 2.5`, that expression is equivalent to
`\left|Y - \mu \right| < 2\, \sigma`.

The percentage of individuals with a height in the interval
`(58.6,\, 68.6)` would be equal to the probability
`P(\left|Y - \mu \right| < 2\, \sigma)`.

Compare this expression to Chebyshev's Theorem. Notice that
`k = 2`. Therefore:

`\displaystyle P\left(\left|Y -\mu\right| \le 2\, \sigma\right) \ge 1 - (1)/(2^2) = 0.75`.

In other words, at least
`75\%` of the heights would be between
`58.6` inches and
`68.6` inches.