Question

Listed below are test grades for certain test.

27, 67, 85, 38, 92, 99, 87, 40, 56, 78, 88, 93, 97, 70, 66,

88, 45, 75, 90, 82, 65, 73, 85, 72, 65, 68, 90, 44, 56, 79 Given that the population mean is 72, and the population standard deviation is 19.

(1) Use the 68 − 95 − 99.7 rule to estimate the percentages of observations that lie with one, two, and three standard deviations of the mean.

(2) Use the data to obtain the exact percentages of observations that lie within one, two, and three standard deviations of the mean.

(3) Is it appropriate to use the 68 − 95 − 99.7 rule in this situation? Why?

(4) Regardless of your conclusion at (3). Suppose the grades of a test follows a Normal distribution with mean 72 and

standard deviation 19.

(a) What percentile are you in if your grade is 85?

(b) If you want to be in the top 5%, what grade do you need?

Answer 1

1)

about 68% of observation of data lie within 1 std dev away from mean
about 95% of observation of data lie within 2 std dev away from mean
about 99.7% of observation of data lie within 3 std dev away from mean

2)

X̄ ± 1 * s = (   53.00   ,   91.00   )

percentage = 21/30 = 70%

X̄ ± 2 * s = (   34.00       110.00   )

percentage=29/30 = 96.67%

X̄ ± 3 * s = (   15.00       129.00   )

percentage=30/30 = 100%

3)

yes, because percentage obtained are approximately the same

4)

a) P( X ≤    85   ) = P( (X-µ)/σ ≤ (85-72) /19)      
=P(Z ≤   0.68   ) =   0.7531  

percentile = 75.31th (answer)

b)

µ=   72                  
σ =    19                  
P(X≤x) =   0.9500                  
                      
Z value at    0.95   =   1.6449   (excel formula =NORMSINV(   0.95   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   1.645   *   19   +   72  
X   =   103.25   (answer)