Answer:
Chebyshev's Theorem (Tchebysheff's Theorem) suggests that this percentage would be greater than [tex]75\%[/tex].
Step-by-step explanation:
Let [tex]Y[/tex] be a random variable with a mean of [tex]\mu[/tex] and a (finite) variance of [tex]\sigma[/tex]. By Chebyshev's Theorem, for any constant [tex]k[/tex] where [tex]k > 0[/tex]:
[tex]\displaystyle P\left(\left|Y -\mu\right| \le k\, \sigma\right) \ge 1 - \frac{1}{k^2}[/tex].
In this question, let [tex]Y[/tex] denote height (in inches.) Accordingly, [tex]\mu = 63.6[/tex] whereas [tex]\sigma = 2.5[/tex].
Note, that the interval of interest [tex](58.6,\, 68.6)[/tex] is centered at [tex]\mu = 63.6[/tex]. Besides, the difference between [tex]\mu[/tex] and either endpoints of this interval is equal to [tex]5.0[/tex], which is the same as [tex]2\, \sigma[/tex].
Rewrite the interval [tex](58.6,\, 68.6)[/tex] as [tex](63.6 - 2\times 2.5,\, 63.6 + 2 \times 2.5)[/tex].
For the height [tex]Y[/tex] of one such individual, [tex]Y \in (58.6,\, 68.6)[/tex] is equivalent to [tex]\left|Y - 63.6\right| < 2 \times 2.5[/tex]. With [tex]\mu = 63.6[/tex] whereas [tex]\sigma = 2.5[/tex], that expression is equivalent to [tex]\left|Y - \mu \right| < 2\, \sigma[/tex].
The percentage of individuals with a height in the interval [tex](58.6,\, 68.6)[/tex] would be equal to the probability [tex]P(\left|Y - \mu \right| < 2\, \sigma)[/tex].
Compare this expression to Chebyshev's Theorem. Notice that [tex]k = 2[/tex]. Therefore:
[tex]\displaystyle P\left(\left|Y -\mu\right| \le 2\, \sigma\right) \ge 1 - \frac{1}{2^2} = 0.75[/tex].
In other words, at least [tex]75\%[/tex] of the heights would be between [tex]58.6[/tex] inches and [tex]68.6[/tex] inches.
