Question

the following data for a random sample of banks in two cities represent the atm fees for using another banks atm. compute the range and sample standard deviation for atm fees for each city. which city has the most dispersion based on range? which city has the most dispersion based on the standard deviation?

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$$\begin{array}{lccccc} \\text{city a} & 1.25 & 1.00 & 1.50 & 1.50 & 1.50 \\\\ \\text{city b} & 2.50 & 1.50 & 1.25 & 0.00 & 1.75 \\end{array}$$

the range for city a is \\$

the range for city b is \\$

the standard deviation for city a is \\$
(round to the nearest cent as needed.)

the standard deviation for city b is \\$
(round to the nearest cent as needed.)

which city has the most dispersion based on range?

a. city a, because it has a higher range
b. city b, because it has a higher range
c. city a, because it has a lower range
d. city b, because it has a lower range

which city has the most dispersion based on standard deviation?

Explanation:

⚡ Using what you learned: Standard Deviation

Step 1: Calculate the range for City A and City B

The range is the difference between the maximum and minimum values in a data set:

$$ \text{Range} = \text{Maximum} - \text{Minimum} $$

For City A:

• Data: \( 1.25, 1.00, 1.50, 1.50, 1.50 \)
• Maximum = \( 1.50 \), Minimum = \( 1.00 \)

$$ \text{Range}_A = 1.50 - 1.00 = 0.50 $$

For City B:

• Data: \( 2.50, 1.50, 1.25, 0.00, 1.75 \)
• Maximum = \( 2.50 \), Minimum = \( 0.00 \)

$$ \text{Range}_B = 2.50 - 0.00 = 2.50 $$

Step 2: Calculate the sample standard deviation for City A

First, find the mean (\( \bar{x}_A \)):

$$ \bar{x}_A = \frac{1.25 + 1.00 + 1.50 + 1.50 + 1.50}{5} = \frac{6.75}{5} = 1.35 $$

Next, calculate the sum of squared deviations from the mean:

$$ \sum (x_i - \bar{x}_A)^2 = (1.25 - 1.35)^2 + (1.00 - 1.35)^2 + 3 \times (1.50 - 1.35)^2 $$

$$ = (-0.10)^2 + (-0.35)^2 + 3 \times (0.15)^2 $$

$$ = 0.0100 + 0.1225 + 3 \times 0.0225 $$

$$ = 0.0100 + 0.1225 + 0.0675 = 0.2000 $$

Calculate the sample standard deviation (\( s_A \)) with \( n - 1 = 4 \):

$$ s_A = \sqrt{\frac{0.2000}{4}} = \sqrt{0.05} \approx 0.2236 \approx 0.22 $$

Step 3: Calculate the sample standard deviation for City B

First, find the mean (\( \bar{x}_B \)):

$$ \bar{x}_B = \frac{2.50 + 1.50 + 1.25 + 0.00 + 1.75}{5} = \frac{7.00}{5} = 1.40 $$

Next, calculate the sum of squared deviations from the mean:

$$ \sum (x_i - \bar{x}_B)^2 = (2.50 - 1.40)^2 + (1.50 - 1.40)^2 + (1.25 - 1.40)^2 + (0.00 - 1.40)^2 + (1.75 - 1.40)^2 $$

$$ = (1.10)^2 + (0.10)^2 + (-0.15)^2 + (-1.40)^2 + (0.35)^2 $$

$$ = 1.2100 + 0.0100 + 0.0225 + 1.9600 + 0.1225 = 3.3250 $$

Calculate the sample standard deviation (\( s_B \)) with \( n - 1 = 4 \):

$$ s_B = \sqrt{\frac{3.3250}{4}} = \sqrt{0.83125} \approx 0.9117 \approx 0.91 $$

Step 4: Compare dispersion

• Based on range: City B has a range of \( 2.50 \), which is larger than City A's range of \( 0.50 \). Therefore, City B has more dispersion based on range.
• Based on standard deviation: City B has a standard deviation of \( 0.91 \), which is larger than City A's standard deviation of \( 0.22 \). Therefore, City B has more dispersion based on standard deviation.

Answer:

• The range for city A is $0.50
• The range for city B is $2.50
• The standard deviation for city A is $0.22
• The standard deviation for city B is $0.91
• Which city has the most dispersion based on range?

B. City B, because it has a higher range

• Which city has the most dispersion based on standard deviation?

City B, because it has a higher standard deviation