Q. 3.213 - Elementary Statistics

Question

Doing Time. According to the Compendium of Federal Justice Statistics, published by the Bureau of Justice Statistics, the meantime served to first release by Federal prisoners is $32.9$ months. Assume the standard deviation of the times served is $17.9$ months. Let $x$ denote time served to first release by a Federal prisoner.

a. Find the standardized version of $x$.

b. Find the mean and standard deviation of the standardized variable.

c. Determine the $z$-scores for prison times served of $81.3$ months and $20.8$ months. Round your answers to two decimal places.

d. Interpret your answers in part (c)

e. Construct a graph similar to Fig. $3.20$ on page $146$ that depicts your results from parts (b) and (c)

Step-by-Step Solution

Answer

(a) The standardized version of $x$ is, $z=\frac{x-32.9}{17.9}$

(b) The mean and median of standardized variable is, $0, 1$

(c) The $z$-score is, $2.70, -0.67$

(d) The first prisoner's $z$-score indicates that his or her prison term was $2.70$ standard deviations off the mean.

The second prisoner's $z$-score indicates that his or her prison time was $0.67$ standard deviations below the mean.

(e) The required graph is below.

Part (a) Step 1: Given information

We are given that,

The mean is, $\mu =32.9$

And the standard deviation is, $\sigma =17.9$

Part (a) Step 2: Simplify

We know that,

The $x$ is the number of years a federal prisoner has been in custody before being released for the first time.

We use the formula to find the standardized version of $x$ is,

  $z=\frac{x-\mu }{\sigma }=\frac{x-32.9}{17.9}$

Part (b) Step 1: Given information

We need to find out the mean and standard deviation of standardized variable.

Part (b) Step 2: Simplify

The mean of the standardized variable is, 

${\mu }_z=\frac{\sum _{}^{}{z}_i}{n}=\frac{\sum _{}^{}\left({\displaystyle \frac{x_i-\mu }{\sigma }}\right)}{n}=\frac{\sum _{}^{}{x}_i-\sum _{}^{}\mu }{\sigma n}=\frac{\sum _{}^{}{x}_i-n\mu }{\sigma n}=\frac{n\mu -n\mu }{\sigma n}=0$

The standard deviation of the standardized variable is, 

${\sigma }_z=\sqrt{\frac{\sum _{}^{}{\left(z_i-{\mu }_z\right)}^2}{n}}=\sqrt{\frac{\sum _{}^{}{\left(z_i-0\right)}^2}{n}}=\sqrt{\frac{\sum _{}^{}{z}_i^2}{n}}=\frac{\sigma }{\sigma }=1$

Part (c) Step 1: Given information

We need to find out $z$-scores.

Part (c) Step 2: Simplify

The $z$-scores of required data is given below,

$$\begin{gathered} z_1=\frac{x_1-\mu }{\sigma }=\frac{81.3-32.9}{17.9}=2.70 \\ {z}_2=\frac{x_2-\mu }{\sigma }=\frac{20.8-32.9}{17.9}=-0.67 \end{gathered}$$

Here $x_1=81.3$ and $x_2=20.8$ are given values.

Part (d) Step 1: Given information

We need to interpret the answer of part (c)

Part (d) Step 2: Simplify

The first prisoner's $z$-score indicates that his or her prison time was $2.70$ standard deviations out of the mean.

The second prisoner's $z$-score indicates that his or her time in prison was $0.67$ standard deviations below the mean.

Part (e) Step 1: Given information

We need to find the graph.

Part (e) Step 2: Explanation

We know that,

For $x$

   $$\begin{gathered} \mu -\sigma =32.9-17.9=15 \\ \mu +\sigma =32.9+17.9=50.8 \\ \mu -2\sigma =32.9-\left(2\times 17.9\right)=-2.9 \\ \mu +2\sigma =32.9+\left(2\times 17.9\right)=68.7 \\ \mu -3\sigma =32.9-\left(3\times 17.9\right)=-20.8 \\ \mu +3\sigma =32.9+\left(3\times 17.9\right)=86.6 \end{gathered}$$

Similarly, we find the above values for $z$

We have constructed a graph to indicate the results from parts (b) and (c).

And the graph is,