Question

Molly and James tracked the average temperature in their city for the past 8 days. They recorded their temperature findings in this list. 74°F , 79°F , 76°F, 85°F, 87°F, 83°F, 86°F, 78°F What is the mean absolute deviation of this data set?

Answer 1

Answer:

Mean = 81°F.

S. dev. = 5°F.

Step-by-step explanation:

When we have a set of data:

{x₁, x₂, ..., xₙ}

The mean can be calculated as:

M = (x₁ + x₂ ... + xₙ)/N

Where N is the number of data points that we have:

in this case the set is:

{ 74°F , 79°F , 76°F, 85°F, 87°F, 83°F, 86°F, 78°F }

So N = 8.

Then the mean is:

M = ( 74°F + 79°F + 76°F + 85°F + 87°F + 83°F + 86°F + 78°F )/8

M = 81°F.

Now, the standard deviation can be calculated as.

Sd =  √ ( (1/N)*∑(xₐ - M)^2)

where the summation is over xₐ, which represents a summation over all the points in the data set.

Then we can write the standard deviation as:

Sd = √(1/8)*√( (74°F - 81°F)^2 + (79°F - 81°F)^2 + (76°F - 81°F)^2 + (85°F - 81°F)^2 + (87°F - 81°F)^2 + (83°F - 81°F)^2 + (86°F - 81°F)^2 + (78°F - 81°F)^2)

Sd = 4.58°F.

That we should round up to 5°F (because our mean has no digits after the decimal point)