Let's start by finding the variances for the ATM fees in each city.
### Step-by-Step Solution:
1. List the ATM fees for each city:
- City A: [tex]\(1.25, 1.00, 1.50, 1.50, 1.00\)[/tex]
- City B: [tex]\(2.25, 1.25, 1.75, 0.00, 1.75\)[/tex]
2. Calculate the sample variance for City A:
- First, find the mean (average) of the ATM fees for City A:
[tex]\[
\text{Mean of City A} = \frac{1.25 + 1.00 + 1.50 + 1.50 + 1.00}{5} = \frac{6.25}{5} = 1.25
\][/tex]
- Next, calculate the squared differences from the mean:
[tex]\[
(1.25 - 1.25)^2 = 0.00
\][/tex]
[tex]\[
(1.00 - 1.25)^2 = 0.0625
\][/tex]
[tex]\[
(1.50 - 1.25)^2 = 0.0625
\][/tex]
[tex]\[
(1.50 - 1.25)^2 = 0.0625
\][/tex]
[tex]\[
(1.00 - 1.25)^2 = 0.0625
\][/tex]
- Sum of squared differences:
[tex]\[
0.00 + 0.0625 + 0.0625 + 0.0625 + 0.0625 = 0.25
\][/tex]
- Sample variance (using [tex]\(n-1\)[/tex] for sample variance):
[tex]\[
\text{Variance of City A} = \frac{0.25}{4} = 0.0625
\][/tex]
- Round to the nearest cent:
[tex]\[
\text{Variance of City A} = 0.06
\][/tex]
3. Calculate the sample variance for City B:
- First, find the mean (average) of the ATM fees for City B:
[tex]\[
\text{Mean of City B} = \frac{2.25 + 1.25 + 1.75 + 0.00 + 1.75}{5} = \frac{7.00}{5} = 1.40
\][/tex]
- Next, calculate the squared differences from the mean:
[tex]\[
(2.25 - 1.40)^2 = 0.85^2 = 0.7225
\][/tex]
[tex]\[
(1.25 - 1.40)^2 = (-0.15)^2 = 0.0225
\][/tex]
[tex]\[
(1.75 - 1.40)^2 = 0.35^2 = 0.1225
\][/tex]
[tex]\[
(0.00 - 1.40)^2 = (-1.40)^2 = 1.96
\][/tex]
[tex]\[
(1.75 - 1.40)^2 = 0.35^2 = 0.1225
\][/tex]
- Sum of squared differences:
[tex]\[
0.7225 + 0.0225 + 0.1225 + 1.96 + 0.1225 = 2.95
\][/tex]
- Sample variance (using [tex]\(n-1\)[/tex] for sample variance):
[tex]\[
\text{Variance of City B} = \frac{2.95}{4} = 0.7375
\][/tex]
- Round to the nearest cent:
[tex]\[
\text{Variance of City B} = 0.74
\][/tex]
### Final Answer:
- The variance for City A is \[tex]$0.06.
- The variance for City B is \$[/tex]0.74.
