Answer to Question #215088 in Statistics and Probability for Arshad Ali Wazir

Question

Q.5 In a Standard Normal Distribution, find:

i. Mean and Standard Deviation

ii. Lower Quartile

iii. Upper quartile

iv. Inter-quartile range

v. Mean Deviation

Answer 1

A standard normal distribution is a normal distribution with mean zero and variance 1.

i.

Mean = 0 and standard deviation = 1

ii. Lower quartile

Lower quartile is the z value corresponding to the lower 25%.

From z-tables or =NORM.S.INV(0.25) excel function, the lower quartile is $-0.6745$

iii. Upper quartile

Lower quartile is the z value corresponding to the lower 75%.

From z-tables or =NORM.S.INV(0.75) excel function, the lower quartile is $0.6745$

iv.Inter-quartile range

Inter-quartile range is the difference between upper and lower quartile.

$IQR=0.6745+0.6745=1.349$

v. Mean deviation

Generally, mean deviation of a normal distribution is as follows,

$\mathbf{E}\left[\left|X-\mu\right|\right] =\int_{-\infty}^{\infty}\left|a-\mu\right|f_{X}(a)da\\$

$=\int_{-\infty}^{\mu}\left(\mu-a\right)f_{X}(a)da+\int_{\mu}^{\infty}\left(a-\mu\right)f_{X}(a)da\\$

$\overset{1}{=}2\int_{\mu}^{\infty}\left(a-\mu\right)f_{X}(a)da\\$

$=2\int_{\mu}^{\infty}\frac{a-\mu}{\sigma\sqrt{2\pi}}e^{-\left(\frac{a-\mu}{\sigma\sqrt{2}}\right)^{2}}da\\$

$\overset{2}{=}\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}be^{-b^{2}}\sigma\sqrt{2}db\\$

$=2\sqrt{\frac{2}{\pi}}\sigma\int_{0}^{\infty}be^{-b^{2}}db\\$

$=2\sqrt{\frac{2}{\pi}}\sigma\left[\frac{e^{-b^{2}}}{-2}\right]_{0}^{\infty}\\$

$=\sqrt{\frac{2}{\pi}}\sigma\left[e^{0}-e^{-\infty}\right]\\$

$=\sqrt{\frac{2}{\pi}}\sigma.$

But since standard deviation is 1 for a standard normal distribution, mean deviation becomes $\sqrt{\frac{2}{\pi}}$