Chapter 6.2, Problem 56E

a.

To determine

To find: The mean and standard deviation of the difference in the number of errors.

Answer to Problem 56E

The mean and standard deviation are -1.1 and 1.513.

Explanation of Solution

Given:

For X:

Population mean $\left(\mu \right)=2.1$

Population standard deviation $\left(\sigma \right)=1.136$

For Y:

Population mean $\left(\mu \right)=1$

Population standard deviation $\left(\sigma \right)=1$

Calculation:

The mean value and standard deviation of the difference between Y and X can be computed as:

  $\begin{array}{c}{\mu }_{Y-X}={\mu }_Y-{\mu }_X\\ =1.0-2.1\\ =-1.1\end{array}$

The average difference between the errors is expected to be -1.1

  $\begin{array}{c}{\sigma }_{Y-X}=\sqrt{{\sigma }^2{}_Y+{\sigma }^2{}_X}\\ =\sqrt{{1.136}^2+{1.0}^2}\\ =\sqrt{2.290}\\ =1.513\end{array}$

The difference that exists between the two errors on an average is expected to be of -1.1 words and it differs on average of 1.5134 words.

b.

To determine

To find: The probability that randomly chosen person is making more word errors than the non-word errors

Answer to Problem 56E

The probability is 0.15.

Explanation of Solution

Calculation:

The possible values of T are:

  $\begin{array}{l}3-0=3\\ 2-0=2\\ 1-0=1\\ 0-0=0\end{array}$

And,

  $\begin{array}{l}1-2=-1\\ 1-3=-2\\ 0-3=-3\\ 0-4=-4\end{array}$

The probabilities can be calculated as:

  $\begin{array}{c}P\left(Y-X=1\right)=0.3\left(0.1\right)+0.2\left(0.2\right)+0.1\left(0.3\right)\\ =0.10\\ P\left(Y-X=2\right)=0.1\left(0.2\right)+0.2\left(0.1\right)\\ =0.04\end{array}$

And,

  $\begin{array}{c}P\left(Y-X=3\right)=0.1\left(0.1\right)\\ =0.01\end{array}$

Thus,

  $\begin{array}{c}P\left(Y>X\right)=P\left(Y-X>0\right)\\ =0.10+0.04+0.01\\ =0.15\end{array}$

Thus, the required probability is 0.15.