Question

a sample of salaries at a distribution center revealed a mean income of $72,000 with a standard deviation of $4,000. data is not normally distributed. at least what percent of the employees earns at least $65,600 but no more than $78,400? what is k? at least what percent of the employees earns between $65,600 and $78,400? as percent

Explanation:

Step1: Calculate the value of \(k\)

The formula for \(k\) (number of standard deviations from the mean) is \(k=\frac{x - \mu}{\sigma}\).
For \(x = 78400\), \(\mu=72000\), \(\sigma = 4000\), \(k_1=\frac{78400 - 72000}{4000}=\frac{6400}{4000}=1.6\).
For \(x = 65600\), \(k_2=\frac{65600 - 72000}{4000}=\frac{- 6400}{4000}=-1.6\). So \(k = 1.6\).

Step2: Apply Chebyshev's theorem

Chebyshev's theorem states that for any number \(k>0\), the proportion of data within \(k\) standard deviations of the mean is at least \(1-\frac{1}{k^{2}}\).
Substitute \(k = 1.6\) into the formula: \(1-\frac{1}{1.6^{2}}=1-\frac{1}{2.56}\).
\(1-\frac{1}{2.56}=\frac{2.56 - 1}{2.56}=\frac{1.56}{2.56}\approx0.61\) or \(61\%\).

Answer:

\(k = 1.6\), at least \(61\%\)