## Chapter 3

## 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.

## Step-by-Step

## 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.