Use the information given in the table to complete the following statement.

Brenda is 50 inches tall. Her [tex]$z$[/tex]-score is [tex]$\square$[/tex]

\begin{tabular}{|c|c|c|}
\hline
Age & Mean & \begin{tabular}{c}
Standard \\
Deviation
\end{tabular} \\
\hline
7 years & 49 inches & 2 inches \\
\hline
\end{tabular}



Answer :

To find Brenda's [tex]$z$[/tex]-score given her height, we need to follow these steps:

1. Identify the given information:
- Brenda's height is 50 inches.
- We are provided with the mean height for 7-year-olds, which is 49 inches.
- The standard deviation of the heights is 2 inches.

2. Recall the formula for the [tex]$z$[/tex]-score:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where:
- [tex]\( X \)[/tex] is the individual data point (Brenda's height).
- [tex]\( \mu \)[/tex] is the mean of the data set.
- [tex]\( \sigma \)[/tex] is the standard deviation of the data set.

3. Substitute the given values into the formula:
[tex]\[ z = \frac{50 - 49}{2} \][/tex]

4. Perform the calculation:
[tex]\[ z = \frac{1}{2} = 0.5 \][/tex]

Therefore, Brenda's [tex]$z$[/tex]-score is 0.5.