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Sagot :
To compute the sample variance for City A's ATM fees data, we follow these steps:
1. List the given ATM fees:
[tex]\[ 1.25, 1.00, 1.50, 1.50, 1.00 \][/tex]
2. Calculate the mean (average) of the data:
[tex]\[ \text{Mean} = \frac{1.25 + 1.00 + 1.50 + 1.50 + 1.00}{5} = \frac{6.25}{5} = 1.25 \][/tex]
3. Find the squared differences from the mean for each data point:
[tex]\[ (1.25 - 1.25)^2 = 0.00 \][/tex]
[tex]\[ (1.00 - 1.25)^2 = (-0.25)^2 = 0.0625 \][/tex]
[tex]\[ (1.50 - 1.25)^2 = (0.25)^2 = 0.0625 \][/tex]
[tex]\[ (1.50 - 1.25)^2 = (0.25)^2 = 0.0625 \][/tex]
[tex]\[ (1.00 - 1.25)^2 = (-0.25)^2 = 0.0625 \][/tex]
4. Sum these squared differences:
[tex]\[ 0.00 + 0.0625 + 0.0625 + 0.0625 + 0.0625 = 0.25 \][/tex]
5. Calculate the sample variance (noting that we divide by [tex]\(n-1 = 5-1 = 4\)[/tex]):
[tex]\[ s^2 = \frac{0.25}{4} = 0.0625 \][/tex]
6. Round the variance to the nearest cent:
[tex]\[ 0.0625 \approx 0.06 \][/tex]
So, the variance for ATM fees in City A is [tex]\(\$0.06\)[/tex].
1. List the given ATM fees:
[tex]\[ 1.25, 1.00, 1.50, 1.50, 1.00 \][/tex]
2. Calculate the mean (average) of the data:
[tex]\[ \text{Mean} = \frac{1.25 + 1.00 + 1.50 + 1.50 + 1.00}{5} = \frac{6.25}{5} = 1.25 \][/tex]
3. Find the squared differences from the mean for each data point:
[tex]\[ (1.25 - 1.25)^2 = 0.00 \][/tex]
[tex]\[ (1.00 - 1.25)^2 = (-0.25)^2 = 0.0625 \][/tex]
[tex]\[ (1.50 - 1.25)^2 = (0.25)^2 = 0.0625 \][/tex]
[tex]\[ (1.50 - 1.25)^2 = (0.25)^2 = 0.0625 \][/tex]
[tex]\[ (1.00 - 1.25)^2 = (-0.25)^2 = 0.0625 \][/tex]
4. Sum these squared differences:
[tex]\[ 0.00 + 0.0625 + 0.0625 + 0.0625 + 0.0625 = 0.25 \][/tex]
5. Calculate the sample variance (noting that we divide by [tex]\(n-1 = 5-1 = 4\)[/tex]):
[tex]\[ s^2 = \frac{0.25}{4} = 0.0625 \][/tex]
6. Round the variance to the nearest cent:
[tex]\[ 0.0625 \approx 0.06 \][/tex]
So, the variance for ATM fees in City A is [tex]\(\$0.06\)[/tex].
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