Answer:
We have been given the data: 47,45,44,41,48
![Mean=\frac{\text{sum of observations}}{\text{number of observations}}](https://img.qammunity.org/2019/formulas/mathematics/middle-school/xvprl6tqokzsygtc8fr1z0gpp2t98sqbuf.png)
On substituting the values in the above formula to find mean:
![Mean=(47+45+44+41+48)/(5)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/8ydfh1oy739jqcvomicotunx7m14y369y9.png)
![Mean=(225)/(5)=45](https://img.qammunity.org/2019/formulas/mathematics/middle-school/8lddimmupo7pc68cz2hxyohll0e9v7rz8x.png)
Now, we need to find the absolute deviation that is:
![\text{absolute deviation}=\sum{|x-\bar{x}|}](https://img.qammunity.org/2019/formulas/mathematics/middle-school/c7b5em227kresf1qiam0zyd93qryq8yb88.png)
Where
is the mean and x is the values given of x which are: 2,0,1,4
![\text{absolute deviation}=|2-45|+|0-45|+|1-45|+|4-45|=43+45+44+41](https://img.qammunity.org/2019/formulas/mathematics/middle-school/pwya29vmfn79ifghc78p5z10uvoq7gqxod.png)
![\Rightarrow \text{absolute deviation}=173](https://img.qammunity.org/2019/formulas/mathematics/middle-school/xvy3nvl84n551otn7fg9rctxgeak1rvnjy.png)
Now, to find mean absolute deviation we have a formula:
![\text{mean absolute deviation}=\sum\frac{(x-\bar{x})}{N}](https://img.qammunity.org/2019/formulas/mathematics/middle-school/klkg1uwidxnukr5gwvfjfnifcmw3l9i5wn.png)
![\text{mean absolute deviation}=(|47-45|+|45-45|+|44-45|+|41-45|+|48-45|)/(5)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/m4k1a3tgql2dk3hilngo4mkr4euskxc7ka.png)
![\text{mean absolute deviation}=(2+0+1+4+3)/(5)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/buh5vqvu9iznirofs9ua3hoj78h4oyufh9.png)
![\text{mean absolute deviation}=2](https://img.qammunity.org/2019/formulas/mathematics/middle-school/53vaxvtdwqp6ree0zewrtqvj2d3rrpdrhv.png)