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NORMAL DISTRIBUTION — Statistics STD 12 Commerce — Question

Gujarat BoardEnglish MediumSTD 12 CommerceStatisticsNORMAL DISTRIBUTION3 Marks
Question
A normal variable $X$ has mean $200$ and variance $100 .$
$(i)$ Estimate the values of extreme quartiles.
$(ii)$ Find the approximate value of quartile deviation.
$(III)$ Find the approximate value of mean deviation.
✓

Answer

Here, $\mu=200, \sigma^2=100,$
$\therefore \sigma=10$
$(i)$ Extreme quartiles $Q_1 Q_3$ :
$25 \%$ of the observations of the distribution is less than $Q_1$ and $25 \%$ of observations of the distribution Is more than $\mathrm{Q}_3$.
For $\mathrm{Q}_1, \mathrm{z}_1=\frac{Q_1-\mu}{\sigma}=\frac{Q_1-200}{10}$ and
For $\mathrm{Q}_3, \mathrm{z}_3=\frac{Q_3-\mu}{\sigma}=\frac{Q_3-200}{10}$
Image
$P\left[X \leq Q_1\right]=P\left[Z \leq z_1\right]=0.25 \text { and }$
$ P\left[X \geq Q_3\right]=P\left[Z \geq z_2\right]=0.25$
$ \therefore P\left[z_1 \leq Z \leq 0\right]=0.50-0.25=0.25 \text { and }$
$ P\left[0 \leq Z \leq z_2\right]=0.50-0.25=0.25$
$ \text { For area } 0.2486, z_1=-0.67 \text { and for area } 0.2518, z_1=-0.68$
$ \text { For area } 0.25, z_1=\frac{(-0.67)+(-0.68)}{2}$
$ =-0.675$
$ \text { and for area } 0.25 \text {. }$
$ z_2=\frac{0.67+0.68}{2}=0.675$
$ \text { Now, } z_1=\frac{Q_1-200}{10}$
$ \therefore-0.675=\frac{Q_1-200}{10}$
$ \therefore-6.75=Q_1-200$
$ \therefore Q_1=200-6.75$
$ \therefore Q_1=193.25$
$ \text { Now, } z_2=\frac{Q_3-200}{10}$
$ \therefore 0.675=\frac{Q_3-200}{10}$
$ \therefore 6.75=\mathrm{Q}_3-200$
$ \therefore Q_3=200+6.75$
$ \therefore Q_3=206.75$
$(ii)$ The approximate value of quartile deviation: In normal distribution, quartile deviation $\approx \frac{2}{3} \sigma$
Putting, $\sigma=10$ Quartile deviation $\approx \frac{2}{3} \times 10=\frac{20}{3} \approx 6.67$
Hence, the approximate value of quartile deviation obtained is $6.67 .$
$​​​​​​​(iii)$ Approximate value of mean deviation: In normal distribution, mean deviation $\approx \frac{4}{5} \sigma$
Putting, $\sigma=10$ Mean deviation $\approx \frac{4}{5} \times 10 \approx 8$
Hence, the approximate value of mean deviation obtained is $8 .$

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