Question

express the amount of data in each section of the standard normal distribution as a percentage (%).
a =
%
b =
%
c =
%
d =
%
e =
%
f =
%
g =
%
h =
%

Explanation:

Step1: Calculate the total percentage outside of \(99.7\%\)

The total percentage under the normal distribution curve is \(100\%\). The percentage within \(- 3\sigma\) to \(3\sigma\) is \(99.7\%\). So the percentage outside of \(-3\sigma\) and \(3\sigma\) (i.e., \(A + H\)) is \(100\% - 99.7\%=0.3\%\). Since the normal distribution is symmetric, \(A = H=\frac{0.3\%}{2}=0.15\%\)

Step2: Calculate the percentage between \(-3\sigma\) and \(-2\sigma\) (i.e., \(B\)) and between \(2\sigma\) and \(3\sigma\) (i.e., \(G\))

The percentage within \(-2\sigma\) to \(2\sigma\) is \(95\%\). The percentage within \(-3\sigma\) to \(3\sigma\) is \(99.7\%\). So the percentage between \(-3\sigma\) and \(-2\sigma\) (and also between \(2\sigma\) and \(3\sigma\)) is \(\frac{99.7\% - 95\%}{2}=\frac{4.7\%}{2} = 2.35\%\). So \(B = G=2.35\%\)

Step3: Calculate the percentage between \(-2\sigma\) and \(-1\sigma\) (i.e., \(C\)) and between \(1\sigma\) and \(2\sigma\) (i.e., \(F\))

The percentage within \(-1\sigma\) to \(1\sigma\) is \(68\%\). The percentage within \(-2\sigma\) to \(2\sigma\) is \(95\%\). So the percentage between \(-2\sigma\) and \(-1\sigma\) (and also between \(1\sigma\) and \(2\sigma\)) is \(\frac{95\% - 68\%}{2}=\frac{27\%}{2}=13.5\%\). So \(C = F = 13.5\%\)

Step4: Calculate the percentage between \(-1\sigma\) and \(0\) (i.e., \(D\)) and between \(0\) and \(1\sigma\) (i.e., \(E\))

Since the percentage within \(-1\sigma\) to \(1\sigma\) is \(68\%\), and the normal distribution is symmetric about \(x = \mu\) (where \(z = 0\)), \(D=E=\frac{68\%}{2}=34\%\)

Answer:

\(A = 0.15\%\)
\(B = 2.35\%\)
\(C = 13.5\%\)
\(D = 34\%\)
\(E = 34\%\)
\(F = 13.5\%\)
\(G = 2.35\%\)
\(H = 0.15\%\)