Answer:
[tex] \bar X_1 =\frac{30+ 38+ 66+ 78+ 47+ 75+ 35+ 45+ 56+ 29+ 49+ 37}{12}=48.75[/tex]
The mean average monthly temperature in City 1 is 48.75°F.
[tex] MAD = \frac{\sum_{i=1}^n |X_i -\bar X|}{n} = \frac{160.5}{12}= 13.375[/tex]
The mean absolute deviation for the average monthly temperature in City 1 is 13.375°F
Step-by-step explanation:
For this case we have the following dataset given:
Average Monthly Temperatures for City 1 (° F)
30, 38, 66, 78, 47, 75, 35, 45, 56, 29, 49, 37
Average Monthly Temperatures for City 2 (° F)
15, 23, 51, 63, 32, 60, 20, 30, 41, 14, 34, 22
For this case the sample mean can be calculated with the following formula:
[tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}[/tex]
And replacing we got:
[tex] \bar X_1 =\frac{30+ 38+ 66+ 78+ 47+ 75+ 35+ 45+ 56+ 29+ 49+ 37}{12}=48.75[/tex]
The mean average monthly temperature in City 1 is 48.75°F.
And now we can calculate the following values:
[tex] |30-48.75| =18.75[/tex]
[tex] |38-48.75| =10.75[/tex]
[tex] |66-48.75| =17.25[/tex]
[tex] |78-48.75| =29.25[/tex]
[tex] |47-48.75| =1.75[/tex]
[tex] |75-48.75| =26.25[/tex]
[tex] |35-48.75| =13.75[/tex]
[tex] |45-48.75| =3.75[/tex]
[tex] |56-48.75| =7.25[/tex]
[tex] |29-48.75| =19.75[/tex]
[tex] |49-48.75| =0.25[/tex]
[tex] |37-48.75| =11.75[/tex]
And the mean absolute deviation is given by:
[tex] MAD = \frac{\sum_{i=1}^n |X_i -\bar X|}{n} = \frac{160.5}{12}= 13.375[/tex]
The mean absolute deviation for the average monthly temperature in City 1 is 13.375°F
