Assume the random variable X is normally distributed, with mean
μ=41 and standard deviation σ=9. Find the 5th percentile.
+2094 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!
+36916 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
