Question

1. the mean of a distribution is 20 and the standard deviation is 2. use chebyshevs theorem.

a. at least what percentage of the values will fall between 10 and 30?
b. at least what percentage of the values will fall between 12 and 28?

Explanation:

Step1: Find the mean

Let the mean be \(\mu = 20\) and the standard deviation \(\sigma=2\).

Step2: For part a

Find \(k\) value. The interval is \(10\) to \(30\). The distance from the mean is \(|x-\mu|\). For \(x = 10\), \(|10 - 20|=10\) and for \(x = 30\), \(|30 - 20| = 10\). Then \(k=\frac{|x-\mu|}{\sigma}=\frac{10}{2}=5\).
By Chebyshev's theorem \(P(|X-\mu|\leq k\sigma)\geq1-\frac{1}{k^{2}}\). Substitute \(k = 5\) into the formula: \(1-\frac{1}{5^{2}}=1-\frac{1}{25}=\frac{24}{25}=0.96 = 96\%\)

Step3: For part b

Find \(k\) value. The interval is \(12\) to \(28\). For \(x = 12\), \(|12 - 20|=8\) and for \(x = 28\), \(|28 - 20| = 8\). Then \(k=\frac{|x-\mu|}{\sigma}=\frac{8}{2}=4\).
By Chebyshev's theorem \(P(|X - \mu|\leq k\sigma)\geq1-\frac{1}{k^{2}}\). Substitute \(k = 4\) into the formula: \(1-\frac{1}{4^{2}}=1-\frac{1}{16}=\frac{15}{16}=0.9375=93.75\%\)

Answer:

a. \(96\%\)
b. \(93.75\%\)