Question

Trisha had the following seven distances (in feet) in the triple jump.
29, 30, 28, 32, 28, 31, 27
What is the approximate mean absolute deviation of her distances? Round the mean and mean absolute deviation to the nearest tenth.

0.1
1.7
1.5
10.3

Answer 1

when you find the mean you add all of them together. Then you divide your answer by how many numbers there are that you added. 

So what I did was add 29+30+28+32+29+31+27= 205
then divide 205 by 7 and you get 29.285

Answer 2

Answer:

Mean= 29.3

and MAD= 1.5

Step-by-step explanation:

The data point is given by:

29, 30, 28, 32, 28, 31, 27

The mean of the data point is the average of the data points i.e. it is the ratio of sum of the data points to the total number of data points.

and Mean absolute deviation i.e. MAD is calculated in following steps:

1)  Calculate the mean.

2) Find the absolute difference of each data points from the mean.

3) find the mean of these absolute difference.

The mean of the data points is:

$Mean=\dfrac{29+30+28+32+28+31+27}{7}\\\\\\Mean=\dfrac{205}{7}\\\\\\Mean=29.3$

Now,

|29-29.3|=0.3

|30-29.3|=0.7

|28-29.3|=1.3

|32-29.3|=2.7

|28-29.3|=1.3

|31-29.3|=1.7

|27-29.3|=2.3

The mean of these absolute difference i.e. MAD is:

$MAD=\dfrac{0.3+0.7+1.3+2.7+1.3+1.7+2.3}{7}\\\\\\MAD=\dfrac{10.3}{7}\\\\\\MAD=1.4714$

which to the nearest tenth is: MAD=1.5