Answer :
Given:
City A: 2.50 1.50 1.25 0.00 2.00
City B: 1.25 1.00 1.50 1.00 1.00
To find dispersion based on range:
The difference between the maximum and minimum values in a set of data is the range.
2.5 is the maximum value of data of city A and 0.00 is the minimum value of data of city A.
Hence, the range of data of city A is,
[tex]R_A=2.50-0.00=2.50[/tex]1.5 is the maximum value of data of city B and 1.00 is the minimum value of data of city B.
Hence, the range of data of city B is,
[tex]R_B=1.50-1.00=0.50[/tex]Since the range of city A is greater than that of city B, city A has most dispersion based on range.
To find dispersion based on the standard deviation:
The data for city A is,
2.50 1.50 1.25 0.00 2.00
The mean of city A is,
[tex]\begin{gathered} \mu_A=\frac{2.5+1.5+1.25+0+2}{5} \\ =1.45 \end{gathered}[/tex]Let each individual value is represented by xi. Then, the squared difference of each individual value of city A is,
[tex](x_i-\mu_A)^2[/tex]Now, find the square of difference of each individual value of city A is,
[tex]\begin{gathered} _{}(2.5-1.45)^2=(1.05)^2=1.1025 \\ (1.5-1.45)^2=(0.05)^2=0.0025 \\ (1.25-1.45)=(-0.2)^2=0.04 \\ (0.00-1.45)^2=(-1.45)^2=2.1025 \\ (2.00-1.45)^2=(0.55)^2=0.3025 \end{gathered}[/tex]The number of values in the data set is n=5.
Let each individual value is represented by xi, then the sample standard deviation is,
[tex]S_A=\frac{1}{n-1}\sum ^n_{i\mathop=1}(x_i-\mu_A)^2[/tex]Hence, the sample standard deviation of city A can be calculated as,
[tex]\begin{gathered} S_A=\sqrt{\frac{1}{5-1}(1.1025+0.0025_{}+0.04+2.1025+0.3025)}_{} \\ =\sqrt[]{\frac{3.55}{4}} \\ =0.9420 \end{gathered}[/tex]Therefore, the sample standard deviation of city A is 0.9420.
The data for city B is:
1.25, 1.00, 1.50, 1.00, 1.00
The mean of city B is,
[tex]\begin{gathered} \mu_B=\frac{1.25+1.00+1.50+1.00+1.00}{5} \\ =1.15 \end{gathered}[/tex]Let each individual value is represented by xi. Then, the squared difference of each individual value of city B is,
[tex]\begin{gathered} (x_i-\mu)^2 \\ (1.25-1.15)^2=0.1^2=0.01 \\ (1.00-1.15)^2=(-0.15)^2=0.0225 \\ (1.5-1.15)^2=(0.35)^2=0.1225 \\ (1.00-1.15)^2=(-0.15)^2=0.0225 \\ (1.00-1.15)^2=(-0.15)^2=0.0225 \end{gathered}[/tex]The number of values in the data set is n=5.
Let each individual value is represented by xi, then the sample standard deviation of city B is,
[tex]S_B=\frac{1}{n-1}\sum ^n_{i\mathop=1}(x_i-\mu_B_{})^2[/tex]Hence, the sample standard deviation of city B can be calculated as,
[tex]\begin{gathered} S_B=\sqrt[]{\frac{1}{5-1}(0.01+0.0225_{}+0.1225+0.0225+0.0225)}_{} \\ =\sqrt[]{\frac{0.2}{4}} \\ =0.2236 \end{gathered}[/tex]Therefore, the sample standard deviation of city B is 0.2236.
Since the standard deviation of city A is greater than that of city B, city A has more dispersion based on the standard deviation.
