## Q. 3.26

Travel Allowances

A survey of local companies found that the mean amount of travel allowance for couriers was $0.25 per mile. The standard deviation was$0.02. Using Chebyshev’s theorem, find the minimum percentage of the data values that will fall between $0.20 and$0.30.

## Verified Solution

Step 1 Subtract the mean from the larger value.

$\ 0.30-\ 0.25=\ 0.05$

Step 2 Divide the difference by the standard deviation to get k.

$k=\frac{0.05}{0.02}=2.5$

Step 3 Use Chebyshev’s theorem to find the percentage.

$1-\frac{1}{k^2}=1-\frac{1}{2.5^2}=1-\frac{1}{6.25}=1-0.16=0.84 \quad \text { or } \quad 84 \%$

Hence, at least 84 % of the data values will fall between $0.20 and$0.30.