Answer:
Approximately [tex](-1.86)[/tex].
Step-by-step explanation:
In a distribution of mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the [tex]z[/tex]-score of an observed value [tex]x[/tex] is:
[tex]\displaystyle z = \frac{x - \mu}{\sigma}[/tex].
In this question:
- The mean of the distribution is [tex]\mu = 69.0[/tex] inches.
- The standard deviation of the distribution is [tex]\sigma = 2.8[/tex] inches.
- The observed value is [tex]x = 63.8[/tex] inches.
Hence, the [tex]z[/tex]-score for the observed value [tex]x = 63.8[/tex] would be:
[tex]\displaystyle z = \frac{x - \mu}{\sigma} = \frac{63.8 - 69.0}{2.8} \approx -1.86[/tex].
