Answer with Step-by-step explanation:
For the tap water the observations are summarized as below:
Tap water sample:
[tex]\bar{pH}_{tap}=\frac{\sum_{i=1}^{N}x_i}{N}[/tex]
where
[tex]x_i[/tex] is the [tex]i^{th}[/tex] observation
'N' is the total number of observations
Applying the given values we get
[tex]\bar{pH}=\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}\\\\\bar{pH}_{tap}=7.5[/tex]
To obtain the median of the given data arranging the given data in ascending order we get
7.10, 7.45,7.45, 7.47,7.47,7.47,7.50, 7.52, 7.56,7.64, 7.68,7.69
since the data is even thus the median of the given data is [tex]Median=\frac{7.47+7.50}{2}=7.485[/tex]
Mode of the data is the value that occurs most frequently thus for the given data mode is 7.47
Part b)
Bottled water sample:
[tex]\bar{pH}_{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}\\\\\bar{pH}_{bottle}=5.194[/tex]
To obtain the median of the given data arranging the given data in ascending order we get
5.02, 5.09,5.13, 5.15,5.20,5.21,5.23, 5.24, 5.26,5.26, 5.26,5.28
since the data is even thus the median of the given data is [tex]Median=\frac{5.21+5.23}{2}=5.422[/tex]
Mode of the data is the value that occurs most frequently thus for the given data mode is 5.26
Part c)
If the 7.10 water measurement is taken incorrectly as 1.70 the mean changes to
[tex]\bar{pH}=\frac{7.64+7.45+7.47+7.50+7.68+7.69+7.45+1.70+7.56+7.47+7.52+7.47}{12}\\\\\bar{pH}_{tap}=7.05[/tex]
