what is the standard deviation for 14,13,6,13,6,14 rounded to one decimal place?

Statistical file:
{14, 13, 6, 13, 6, 14}


Average (mean): μ=11
Absolute deviation: 20
Mean deviation: 3.33333333333
Minimum: 6
Maximum: 14
Variance: 12.6666666667
Standard deviation σ=3.55902608401
Corrected sample standard deviation s=3.89871773792
Coefficient of variation cV=0.354428885266
Mode: {6, 13, 14} - multimodal
Geometric mean: 10.2977152692
Harmonic mean: 9.52325581395
Sum: 66
Sum of absolute values: 66
Average absolute deviation: 3.33333333333
Range: 8

I dont know

Find the (sample) standard deviation of the list:
(14, 13, 6, 13, 6, 14)

The standard deviation is given by:
sqrt((variance))

The (sample) variance of a list of numbers X = {X_1, X_2, ..., X_n} with mean μ = (X_1+X_2+...+X_n)/n is given by:
(abs(X_1-μ)^2+abs(X_2-μ)^2+...+abs(X_n-μ)^2)/(n-1)

There are n = 6 elements in the list X = {14, 13, 6, 13, 6, 14}:
(abs(X_1-μ)^2+abs(X_2-μ)^2+abs(X_3-μ)^2+abs(X_4-μ)^2+abs(X_5-μ)^2+abs(X_6-μ)^2)/(6-1)

The elements X_i of the list X = {14, 13, 6, 13, 6, 14} are:
X_1 = 14
X_2 = 13
X_3 = 6
X_4 = 13
X_5 = 6
X_6 = 14
(abs(14-μ)^2+abs(14-μ)^2+abs(13-μ)^2+abs(13-μ)^2+abs(6-μ)^2+abs(6-μ)^2)/(6-1)

The mean (μ) is given by

μ = (X_1+X_2+X_3+X_4+X_5+X_6)/6 = (14+13+6+13+6+14)/6 = 11:
(abs(14-11)^2+abs(13-11)^2+abs(6-11)^2+abs(13-11)^2+abs(6-11)^2+abs(14-11)^2)/(6-1)

The values of the differences are:
14-11 = 3
13-11 = 2
6-11 = -5
13-11 = 2
6-11 = -5
14-11 = 3
6-1 = 5
(abs(3)^2+abs(2)^2+abs(-5)^2+abs(2)^2+abs(-5)^2+abs(3)^2)/5

The values of the terms in the numerator are:
abs(3)^2 = 9
abs(2)^2 = 4
abs(-5)^2 = 25
abs(2)^2 = 4
abs(-5)^2 = 25
abs(3)^2 = 9
(9+4+25+4+25+9)/5

9+4+25+4+25+9 = 76:
76/5

The standard deviation is given by
sqrt((variance)) = sqrt(76/5) = 2 sqrt(19/5):
Answer: |2 sqrt(19/5)