Questions: The blowyy drits for a random sample of banks in two cities represent the ATM fees for using another bank's ATM. Compute the sample variance for ATM fees for each city ChyA 225 100 1.50 000 175 Chy 8 125 100 150 125 125 The variance for city A is 5073. (Round to the nearest cent as needed) The variance for city B is . (Round to the nearest cent as needed)

Solution Steps

To compute the sample variance for ATM fees for each city, we need to follow these steps:

1. Calculate the mean (average) of the ATM fees for each city.
2. Subtract the mean from each ATM fee to find the deviation of each fee from the mean.
3. Square each deviation.
4. Sum all the squared deviations.
5. Divide the sum by the number of data points minus one (n-1) to get the sample variance.

For City A: \[ \text{Mean}_{A} = \frac{2.25 + 1.00 + 1.50 + 0.00 + 1.75}{5} = \frac{6.50}{5} = 1.30 \]

For City B: \[ \text{Mean}_{B} = \frac{1.25 + 1.00 + 1.50 + 1.25 + 1.25}{5} = \frac{6.25}{5} = 1.25 \]

For City A: \[ \begin{align_} (2.25 - 1.30)^2 & = 0.9025 \\ (1.00 - 1.30)^2 & = 0.0900 \\ (1.50 - 1.30)^2 & = 0.0400 \\ (0.00 - 1.30)^2 & = 1.6900 \\ (1.75 - 1.30)^2 & = 0.2025 \\ \end{align_} \]

For City B: \[ \begin{align_} (1.25 - 1.25)^2 & = 0.0000 \\ (1.00 - 1.25)^2 & = 0.0625 \\ (1.50 - 1.25)^2 & = 0.0625 \\ (1.25 - 1.25)^2 & = 0.0000 \\ (1.25 - 1.25)^2 & = 0.0000 \\ \end{align_} \]

For City A: \[ \text{Sum of Squared Deviations}_{A} = 0.9025 + 0.0900 + 0.0400 + 1.6900 + 0.2025 = 2.9250 \]

For City B: \[ \text{Sum of Squared Deviations}_{B} = 0.0000 + 0.0625 + 0.0625 + 0.0000 + 0.0000 = 0.1250 \]

For City A: \[ \text{Variance}_{A} = \frac{2.9250}{5 - 1} = \frac{2.9250}{4} = 0.7313 \]

For City B: \[ \text{Variance}_{B} = \frac{0.1250}{5 - 1} = \frac{0.1250}{4} = 0.0313 \]

Final Answer