Chapter 2, Problem T2.3SPT

To determine

To Calculate: the mean and standard deviation of the corrected pH measurements.

Answer to Problem T2.3SPT

5.64, 1.32

Explanation of Solution

Given:

  $\begin{array}{l}{\mu }_x=4.60\\ {\sigma }_x=1.10\end{array}$

Formula used:

Properties of mean and standard deviation

  $\begin{array}{c}{\mu }_{aX+b}=a{\mu }_X+b\\ {\sigma }_{aX+b}=a{\sigma }_X\end{array}$

Calculation:

Adding 0.1 to every value and then multiply this by 1.2

  $\text{1}\text{.2}\left(x+0.1\right)=1.2x+0.12$

  $\begin{array}{c}{\mu }_{1.2X+0.12}=1.2\left(4.60\right)+0.12\\ =5.64\\ {\sigma }_{1.2X+0.12}=1.2\left(1.10\right)\\ =1.32\end{array}$

Hence, the correct option is (b)