The mean mu of a distribution is 20 and the standard deviati

The mean mu of a distribution is 20 and the standard deviation a is 2. Use Chebyshev\'s theorem to answer the following: At least what percentage of the values will fall between 10 and 30? At least what percentage of the values will fall between 12 and 28?

Solution

A distribution has Mean = 20 and Standard deviation(k) = 2

At least 11/k2 of the distribution\'s values are within k standard deviations of the meanat least 11/k2 of the distribution\'s values are within k standard deviations of the mean. Hence for

(a) The value of k is given by following equation.

20 - 2K = 10 or 20+2K=30

Therefore, K=5   

Hence, the percentage that will falll between 10 and 30 is [1-(1/k2)]*100 = 96%

(b) The value of k is given by following equation.

20 - 2K = 12 or 20+2K=28

Therefore, K=4   

Hence, the percentage that will falll between 12 and 28 is [1-(1/k2)]*100 = 93.75%