Chapter 14.2, Problem 30E

(a)

To determine

To find: The place must a one-gram weight be hung so that the rod will balance at the $50$ -centimetre mark.

Answer to Problem 30E

The weight should be hung $68\text{cm}$ from the end.

Explanation of Solution

Given:

The data that represents distance (in cm) from one end is as:

  $5,20,37,44,52,68,71,85$.

The mean is 50.

Let $x$ be the distance we want so that the rod will balance at the $50$ -centimetre mark.

  $\begin{array}{c}{\displaystyle {\sum }_1^9\left(\text{mean}-f_i\right)}=0\\ \left[\begin{array}{c}\left(50-5\right)+\left(50-20\right)+\left(50-37\right)+\left(50-44\right)+\\ \left(50-52\right)+\left(50-68\right)+\left(50-71\right)+\left(50-85\right)+\left(50-x\right)\end{array}\right]=0\\ 68-x=0\\ x=68\end{array}$

The weight should be hung $68\text{cm}$ from the end.

(b)

To determine

To find: The place must a two-gram weight be hung so that the rod will balance at the $50$ -centimetre mark.

Answer to Problem 30E

The weight should be hung $59\text{cm}$ from the end.

Explanation of Solution

Given:

The data that represents distance (in cm) from one end is as:

  $5,20,37,44,52,68,71,85$.

The mean is 50.

Let $x$ be the distance we want so that the rod will balance at the $50$ -centimetre mark.

  $\begin{array}{c}{\displaystyle {\sum }_1^9\left(\text{mean}-f_i\right)}=0\\ \left[\begin{array}{c}\left(50-5\right)+\left(50-20\right)+\left(50-37\right)+\left(50-44\right)+\\ \left(50-52\right)+\left(50-68\right)+\left(50-71\right)+\left(50-85\right)+\left(50-x\right)+\left(50-x\right)\end{array}\right]=0\\ 118-2x=0\\ x=59\end{array}$

The weight should be hung $59\text{cm}$ from the end.