Question

The data show the average monthly temperatures for two cities over a 6-month period. City 1: {20, 24, 40, 63, 76, 89} City 2: {41, 50, 58, 62, 72, 83} What does the mean absolute deviation (MAD) of the data sets tell you about the temperatures over the 6-month period? Select from the drop-down menus to correctly answer the question. The MAD for City 2 is _______ the MAD for City 1, which means the average monthly temperatures of City 2 vary _______ the average monthly temperatures for City 1.

Answer 1

Answer:

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

Step-by-step explanation:

For comparing the mean absolute deviations of both data sets we have to calculate the mean absolute deviation for both data sets first,

So for city 1:

$Mean = x1 = \frac{20+24+40+63+76+89}{6}$

$x1 = \frac{312}{6}$

$x1 = 52$

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored as the deviation is the distance of value from the mean and it cannot be negative. For this purpose absolute is used)

$20-52 = -32=32\\24-52=-28=28\\40-52=-12=12\\63-52=11\\76-52=24\\89-52=37$

The deviations will be added then.$Mean Absolute Deviation = \frac{32+28+12+11+24+37}{6} \\=\frac{144}{6}\\=24$

So the mean absolute deviation for city 1 is 24 ..

For city 2:

$Mean = x2 = \frac{41+50+58+62+72+83}{6}$

$x2 = \frac{366}{6}$

$x2 = 61$

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored)

$41-61=-20=20\\50-61=-11=11\\58-61=-3=3\\62-61=1\\72-61=11\\83-61=22$

The deviations will be added then.$Mean Absolute Deviation = \frac{20+11+3+1+11+22}{6} \\=\frac{68}{6}\\=11.33$

So the MAD for city 2 is 11.33 ..

So,

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

Answer 2

Answer:

The MAD for City 2 is less than the MAD for City 1, which means the average monthly temperatures of City 2 vary less than the average monthly temperatures for City 1.

Step-by-step explanation:

The given data sets are

City 1 : {20, 24, 40, 63, 76, 89}

City 2 : {41, 50, 58, 62, 72, 83}

Formula for mean:

$Mean=\frac{\sum x}{n}$

Mean of city 1 is

$Mean=\frac{20+24+40+63+76+89}{6}=52$

Mean of city 2 is

$Mean=\frac{41+50+58+62+72+83}{6}=61$

Formula for mean absolute deviation:

$MAD=\dfrac{\sum |x-\mu|}{n}$

where, μ is the mean of data.

MAD for City 1:

$MAD=\dfrac{|20-52|+|24-52|+|40-52|+|63-52|+|76-52|+|89-52|}{6}$

$MAD=24$

MAD for City 2

$MAD=\frac{\left|41-61\right|+\left|50-61\right|+\left|58-61\right|+\left|62-61\right|+\left|72-61\right|+\left|83-61\right|}{6}$

$MAD=\approx 11.33$

From the above calculation we get 24 > 11.33 or MAD of city 1 > MAD of city 2.

The MAD for City 2 is less than the MAD for City 1, which means the average monthly temperatures of City 2 vary less than the average monthly temperatures for City 1.