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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?


User Mario A
by
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1 Answer

1 vote

Answer:


\mu = 23.33, \sigma =6.13

And for this case we are analyzing the value os 33.44 and we can use the z score formula given by:


z=(X -\mu)/(\sigma)

And replacing we got:


z=(33.44 -23.33)/(6.13)= 1.649

We know that the proportion of area beyond the Z score associated with this age is .05 so then the percentile would be: 95

Explanation:

For this case we have the following parameters:


\mu = 23.33, \sigma =6.13

And for this case we are analyzing the value os 33.44 and we can use the z score formula given by:


z=(X -\mu)/(\sigma)

And replacing we got:


z=(33.44 -23.33)/(6.13)= 1.649

We know that the proportion of area beyond the Z score associated with this age is .05 so then the percentile would be: 95

User Citronex
by
7.2k points

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