91Ó°ÊÓ

$$\begin{gathered} Mean=\frac{Sumofallobservations}{Numberofobservations} \\ =\frac{\left(1.72+2.50+2.16+2.13+1.06+2.24+2.31+ \\ 2.03+1.09+1.40+2.57+2.64+1.26+2.05+ \\ 1.19+2.13+1.27+1.51+2.41+1.95\right)}{20} \\ =\frac{37.62}{20} \\ =1.881 \end{gathered}$$

The mean is 1.881. It indicates that, on average, the roughness of the pipes is around 1.881 micrometers.

To calculate the median we will first arrange the data in ascending order,
(1.06, 1.09, 1.19, 1.26, 1.27, 1.40, 1.51, 1.64, 1.72, 1.95, 2.03, 2.05, 2.13, 2.16, 2.24, 2.31, 2.41, 2.50, 2.57, 2.64)

Now, as the sample size is even,

$$\begin{gathered} Median=\frac{Sumoftwovaluesinthecentre}{2} \\ =\frac{1.95+2.03}{2} \\ =\frac{3.98}{2} \\ =1.99 \end{gathered}$$

Therefore, the median is 1.99 micrometer, the roughness of 10 pipes is less than 1.99, and the roughness of the other 10 pipes is more than 1.99.

Mean is the better measure of the central tendency to describe the surface roughness of pipes because it takes into consideration the roughness of all the pipes. Whereas the median only takes into consideration the middle values of the data set.