Question

the following data for a random sample of banks in two cities represent the atm fees for using another banks atm. compute the sample variance for atm fees for each city
city a 1.00 1.00 1.50 1.00 1.25
city b 2.25 1.25 1.25 0.00 1.75
the variance for city a is $0.05
(round to the nearest cent as needed.)
the variance for city b is $
(round to the nearest cent as needed.)

Explanation:

Step1: Calculate the mean of city B data

Let the data for city B be $x_1 = 2.25$, $x_2=1.25$, $x_3 = 1.25$, $x_4=0.00$, $x_5 = 1.75$. The sample size $n = 5$. The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{2.25 + 1.25+1.25 + 0.00+1.75}{5}=\frac{6.5}{5}=1.3$.

Step2: Calculate the squared - differences

$(x_1-\bar{x})^2=(2.25 - 1.3)^2=(0.95)^2 = 0.9025$, $(x_2-\bar{x})^2=(1.25 - 1.3)^2=(- 0.05)^2=0.0025$, $(x_3-\bar{x})^2=(1.25 - 1.3)^2=(-0.05)^2 = 0.0025$, $(x_4-\bar{x})^2=(0.00 - 1.3)^2=(-1.3)^2 = 1.69$, $(x_5-\bar{x})^2=(1.75 - 1.3)^2=(0.45)^2=0.2025$.

Step3: Calculate the sample variance

The sample variance formula is $s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}$. $\sum_{i = 1}^{n}(x_i-\bar{x})^2=0.9025+0.0025 + 0.0025+1.69+0.2025=2.8025$. Then $s^2=\frac{2.8025}{4}=0.700625\approx0.70$.

Answer:

$0.70$