Answer:
[tex] \mu = 23.33, \sigma =6.13[/tex]
And for this case we are analyzing the value os 33.44 and we can use the z score formula given by:
[tex] z=\frac{X -\mu}{\sigma}[/tex]
And replacing we got:
[tex] z=\frac{33.44 -23.33}{6.13}= 1.649[/tex]
We know that the proportion of area beyond the Z score associated with this age is .05 so then the percentile would be: 95
Step-by-step explanation:
For this case we have the following parameters:
[tex] \mu = 23.33, \sigma =6.13[/tex]
And for this case we are analyzing the value os 33.44 and we can use the z score formula given by:
[tex] z=\frac{X -\mu}{\sigma}[/tex]
And replacing we got:
[tex] z=\frac{33.44 -23.33}{6.13}= 1.649[/tex]
We know that the proportion of area beyond the Z score associated with this age is .05 so then the percentile would be: 95