Question

A variable is normally distributed with mean 16 and standard deviation 2.

A. determine the quartiles of the variable obtain and interpret the 85th percentile

B. Find the value that 65% of all possible values of the variable exceed

C. Find the two values that divide the area under the corresponding normal curve into the middle area of 0.95 and two outside areas of 0.025

I see that this question has been answered in the past, however, they used (0.68) * 2 + 16 instead of (0.67)*2+16 when I believe the value is 0.68 and so forth.

Answer 1

Solution:-
Given that mean = 16, sd = 2

A. The quartiles are 25%, 50% and 75% which equate to "areas" under the Standard Normal curve of 0.25, 0.50 and 0.75, respectively.

The z-value for 25% is -0.674
The z-value for 50% is 0
The z-value for 75% is 0.674

Now, a z-value is simply the "number" of standard deviations. Use that fact to find the solutions ...

25% quartile: 16 + (-0.674*2) = 14.652
50% quartile: 16 + (0*2) = 16
75% quartile: 16 + (0.674*2) = 17.348

=> z-value with 0.85 area is 1.036

85th percentile = 16 + (1.036*2) = 18.072

So, 85% of the data falls below 18.072

B. If 65% exceed, then 35% are "below" so find the z-value for 35%

z = -0.385

65% exceed = 16 + (-0.385*2) = 15.23

C. This can be solved using the "Empirical Rule" (check your textbook for details) ...

95% of the data will fall within "2 standard deviations" ...

16 ± (2*2) = 12 to 20