Answer
Tap water pH has more dispersion.
Work Step by Step
$s=\sqrt {\frac{Σ(x_i-x̅)^2}{n-1}}$
Tap:
First, the mean: $x̅_{Tap}=\frac{7.64+7.45+7.47+7.50+7.68+7.69+7.45+7.10+7.56+7.47+7.52+7.47}{12}=7.50$
Now, the standard deviation:
$s_{Tap}=\sqrt {\frac{(7.64-7.50)^2+(7.45-7.50)^2+(7.47-7.50)^2+(7.50-7.50)^2+(7.68-7.50)^2+(7.69-7.50)^2+(7.45-7.50)^2+(7.10-7.50)^2+(7.56-7.50)^2+(7.47-7.50)^2+(7.52-7.50)^2+(7.47-7.50)^2}{12-1}}=0.154$
Bottled:
First, the mean:
$x̅_{Bottled}=\frac{5.15+5.09+5.26+5.20+5.02+5.23+5.28+5.26+5.13+5.26+5.21+5.24}{12}=5.194$
Now, the standard deviation:
$s_{Bottled}=\sqrt {\frac{(5.15-5.194)^2+(5.09-5.194)^2+(5.26-5.194)^2+(5.20-5.194)^2+(5.02-5.194)^2+(5.23-5.194)^2+(5.28-5.194)^2+(5.26-5.194)^2+(5.13-5.194)^2+(5.26-5.194)^2+(5.21-5.194)^2+(5.24-5.194)^2}{12-1}}=0.0805$
$s_{Tap}\gt s_{Bottled}$
