A survey of college students was conducted during final exam week to assess the number of...
A survey of college students was conducted during final exam week to assess the number of hours spent studying each day. The mean number of hours was 5 with a standard deviation of 1.5 hours. The distribution was normal.
How many hours would an individual at the 60th percentile rank study?
Homework Answers
Ans:
Given that
Normal distribution with mean=5 and standard deviation=1.5
For 60th percentile:
P(Z<=z)=0.6
z=normsinv(0.6)=0.2533
P60=5+0.2533*1.5=5.38 hrs
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