Answer: The mean absolute deviation of this data is 4.13 inches.
Step-by-step explanation:
Let X be the set of the heights of the 11 plants.
Then mean of the given data [tex]\bar{x}[/tex]= [tex]\frac{\sum_{i=1}^{n}x_i}{n}\\=\frac{9+4+10+9+5+2+22+10+3+3+5}{11}=7.45\ \text{inches}[/tex]
Now make table 1 , thus from table 1 we have ,
[tex]\\\sum_{i=1}^{n}|x-x_i|}=45.45[/tex]
Mean absolute deviation =[tex]\frac{\sum_{i=1}^{n}|x_i-\bar{x}|}{n}}= [/tex]
[tex]=\frac{45.45}{11}=4.13\ \text{inches}[/tex]
The mean absolute deviation of this data is 4.13 inches.
Answer:
Step-by-step 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.
