Correct Answer - Option 3 :
\(\frac{4}{5}\sigma\)
Normal distribution:
The probability density function of a normal distribution is given by
\(f\left( x \right) = \frac{1}{{\sigma \sqrt {2\pi } }}{e^{ - \frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}}}\)
Where σ is standard deviation
μ is the mean
Mean (a) expression of is define as
\(E\left( x \right) = \;\mathop \smallint \limits_{ - \infty }^\infty x\;f\left( x \right)\;dx\)
Calculation:
Mean deviation about mean μ is,
\(E\left( {\left| {x - \mu } \right|} \right) = \mathop \smallint \limits_{ - \infty }^\infty \left| {x - \mu } \right|f\left( x \right)\;dx\)
\(= \mathop \smallint \limits_{ - \infty }^\infty \left| {x - \mu } \right|\frac{1}{{\sigma \sqrt {2\pi } }}\;{e^{ - \frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}}}\)
\(= \frac{1}{{\sigma \sqrt {2\pi } }}\;\mathop \smallint \limits_{ - \infty }^\infty \left| {x - \mu } \right|{e^{ - \frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}}}dx\)
Let \(\frac{{x - \mu }}{\sigma } = t\)
⇒ x – μ = σt
⇒ dx = σ dt
Now, \(E\left( {\left| {x - \mu } \right|} \right) = \frac{1}{{\sigma \sqrt {2\pi } }}\;\mathop \smallint \limits_{ - \infty }^\infty \sigma \left| t \right|{e^{ - \frac{{{t^2}}}{2}}}\sigma \;dt\)
\(= \frac{\sigma }{{\sqrt {2\pi } }}\;\mathop \smallint \limits_{ - \infty }^\infty \left| t \right|{e^{ - \frac{{{t^2}}}{2}}}dt\)
\(= \frac{{2\sigma }}{{\sqrt {2\pi } }}\;\mathop \smallint \limits_0^\infty t\;{e^{ - \frac{{{t^2}}}{2}}}dt\)
Let \(\frac{{{t^2}}}{2} = z\)
⇒ 2t dt = 2 dz ⇒ t dt = dz
\(= \frac{{\sqrt 2 \;\sigma }}{{\sqrt \pi }}\;\mathop \smallint \limits_0^\infty {e^{ - z}}\;dz = \sqrt {\frac{2}{\pi }} \sigma \left[ { - {e^{ - z}}} \right]_0^\infty \)
\(= \sqrt {\frac{2}{\pi }} \sigma = 0.8\;\sigma = \frac{4}{5}\sigma\)
