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Sagot :
Answer:
The mean average monthly temperature in City 1 is 48.75°F.
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]$|38-48.75|=10.75$[/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]$M A D=\frac{\sum_{i=1}^{n}\left|X_{i}-\bar{X}\right|}{n}=\frac{160.5}{12}=13.375$[/tex]
The mean absolute deviation for the average monthly temperature in City 1 is 13.375°F
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