Using Chebyshev's Theorem, it is found that:
a) The minimum percentage is 75%.
b) The minimum percentage is 84%.
c) The minimum percentage is 89%.
By Chebyshev's Theorem, the minimum percentage of measures within k standard deviations of the mean is:
[tex]P = 100\left(1 - \frac{1}{k^{2}}\right)[/tex]
Item a:
Within 2 standard deviations, hence k = 2, and:
[tex]P = 100\left(1 - \frac{1}{2^{2}}\right) = 100\left(\frac{3}{4}\right) = 75[/tex]
The minimum percentage is 75%.
Item b:
2.5 standard deviations, hence k = 2.5, and:
[tex]P = 100\left(1 - \frac{1}{2.5^{2}}\right) = 84[/tex]
The minimum percentage is 84%.
Item c:
3 standard deviations, hence k = 3, and:
[tex]P = 100\left(1 - \frac{1}{3^{2}}\right) = 89[/tex]
The minimum percentage is 89%.
A similar problem is given at https://brainly.com/question/15050238