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Assume the random variable X is normally​ distributed, with mean

μ=41 and standard deviation σ=9. Find the 5th percentile.

 Mar 23, 2020
 #1
avatar+1968 
+1

I'll do the 14th percentile as an example:

 

Using a standard normal table, the z value that corresponds to an are of 0.14 is -1.08 or z = -1.08

 

The corresponding x value is

 

x = z*σ + µ = -1.08*5 + 41 = -5.4 + 41 = 35.6

 

Now use this to find the 5th percentile. Hope this helped!

 Mar 23, 2020
 #2
avatar+23646 
+1

Look at your negative z-score table for .0500 (or closest to it) for 5 %

 

z = -1.65     and   - 1.64    are equally apart from .0500

 

I'll take the average of these two (you may have been instructed to just choose one)

 

-1.645   standard deviations from the mean of 41

-1.645 ( 9) + 41 = 26.195  is 5th percentile

 Mar 23, 2020

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