Question

Question:

The mean score on an accounting test is 60, with a standard deviation of 8. Between which two scores must this mean lie to represent 8/9 of the data set?

Find first the value of k.

Use Chebychev’s Inequality to solve this.

Answer 1

Mean = 60

Standard deviation = 8

The value of k as per Chebychev’s Inequality rule:

$1-\frac{1}{k^2}=\frac{8}{9}$

$1-\frac{8}{9}=\frac{1}{k^2}$

$\frac{1}{9}=\frac{1}{k^2}$

$k^2=9$

$k=3$

Lower score limit:

$60-(k)(8)=60-(3\times8)=36$

Upper score limit:

$60+(k)(8)=60+(3\times8)=84$

The mean score on an accounting test is 60, with an standard deviation of 8 must lie between 36 and 84 to represent 8/9 of the data set.