Question

Josie recorded the average monthly temperatures for two cities in the state where she lives. For City 1,

Answer 1

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:

$\bar{X}=\frac{\sum_{i=1}^{n} x_{i}}{n}$

And replacing we got:

$\bar{X}_{1}=\frac{30+38+66+78+47+75+35+45+56+29+49+37}{12}=48.75$

The mean average monthly temperature in City 1 is 48.75°F.

And now we can calculate the following values:

$|38-48.75|=10.75$

$|66-48.75|=17.25$

$|78-48.75|=29.25$

$|47-48.75|=1.75$

$|75-48.75|=26.25$

$|35-48.75|=13.75$

$|45-48.75|=3.75$

$|56-48.75|=7.25$

$|29-48.75|=19.75$

$|49-48.75|=0.25$

$|37-48.75|=11.75$

And the mean absolute deviation is given by:

$M A D=\frac{\sum_{i=1}^{n}\left|X_{i}-\bar{X}\right|}{n}=\frac{160.5}{12}=13.375$

The mean absolute deviation for the average monthly temperature in City 1 is  13.375°F