7. The mean age at first marriage for respondents in a survey is 23.33,
with a standard deviation of 6.13. For an age at first marriage of 33.44,
the proportion of area beyond the Z score associated with this age is
.05. What is the percentile rank for this score?

Respuesta :

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