Answer: mean = 10.3 and median = 9.5
Step-by-step explanation:
Given data :
22, 14, 8, 1, 9, 0, 31, 2, 13, 3, 11, 10
Mean = ![\dfrac{\text{Sum of all data values}}{\text{No. of data values}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7BSum%20of%20all%20data%20values%7D%7D%7B%5Ctext%7BNo.%20of%20data%20values%7D%7D)
![=\dfrac{22+14+8+1+9+0+31+2+13+3+11+10}{12}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B22%2B14%2B8%2B1%2B9%2B0%2B31%2B2%2B13%2B3%2B11%2B10%7D%7B12%7D)
![=\dfrac{124}{12}=10.3](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B124%7D%7B12%7D%3D10.3)
Median = Middle most value.
For Median , we first arrange the data values in an order.
0 , 1 , 2, 3, 8, 9, 10 , 11, 13 , 14, 22, 31
Since , number of data values is 12 ( even) , so the median would be the mean of the two middlemost value.
i.e . Median= ![\dfrac{9+10}{2}=9.5](https://tex.z-dn.net/?f=%5Cdfrac%7B9%2B10%7D%7B2%7D%3D9.5)
Hence, the mean and median of this distribution (in years) are :
mean = 10.3 and median = 9.5