Question

(08.06)The following data show the height, in inches, of 11 different plants in a garden: 9 4 10 9 5 2 22 10 3 3 5 After removing the outlier, what does the mean absolute deviation of this data set represent?

Answer 1

Explanation:

Given the data

9, 4, 10, 9, 5, 2, 22, 10, 3, 3, 5

Mean deviation can be calculated using

X = Σ|x − μ| / N.

Where x is the given data

μ is the mean of the data

N is the frequency of data

Step 1: Find the mean:

μ = Σx. / N

μ = 9+4+10+9+5+2+22+10+3+3+5 /11

μ = 82 / 11

μ =7.45

Step 2: Find the distance of each value from that mean:

Note that, |x-μ| means that the magnitude of the value inside the parenthesis and it means it wonts return a negative value, so if the answer is negative, we will write positive

x. |x-μ|

9. 1.55

4. 3.45

10. 2.55

9. 1.55

5. 2.45

2. 5.45

22. 14.55

10. 2.55

3. 4.45

3. 4.45

5. 2.45

Then,

Step 3. Find the mean of those distances

X = Σ|x − μ| / N.

X = 1.55 + 3.45 + 2.55 + 1.55 + 2.45 + 5.45 + 14.55 + 2.55 + 4.55 + 4.55 + 2.45 / 11

X = 45.45 / 11

X = 4.132

So, the mean = 7.45, and the mean deviation = 4.132.

It tells us how far, on average, all values are from the middle.

In that example the values are, on average, 4.132 away from the middle.

Answer 2

Answer: The mean absolute deviation of this data is 4.13 inches.

Explanation:

Let X be the set of the heights of the 11 plants.

Then mean of the given data
`\bar{x}`=
`(\sum_(i=1)^(n)x_i)/(n)\\=(9+4+10+9+5+2+22+10+3+3+5)/(11)=7.45\ \text{inches}`

Now make table 1 , thus from table 1 we have ,

`\\\sum_(i=1)^(n)|x-x_i|}=45.45`

Mean absolute deviation =
`\frac{\sum_(i=1)^(n)|x_i-\bar{x}|}{n}}=`

`=(45.45)/(11)=4.13\ \text{inches}`

The mean absolute deviation of this data is 4.13 inches.