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NORMAL DISTRIBUTION — Statistics STD 12 Commerce — Question

Gujarat BoardEnglish MediumSTD 12 CommerceStatisticsNORMAL DISTRIBUTION3 Marks
Question
For a normal distribution, the first quartile is $35$ and its third quartile is $65 .$ Estimate the limits that includes exactly middle $95.45 \%$ of the observations.
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Answer

Here, $\mathrm{Q}_1=35 ; \mathrm{Q}_3=65$
$\therefore \mathrm{M}=\frac{Q_3+Q_1}{2}=\frac{65+35}{2}=\frac{100}{2}=50$
$ \text { and quartile deviation }=\frac{Q_3-Q_1}{2}$
$ =\frac{65-35}{2}$
$ =\frac{30}{2}=15$
Now, in normal distribution $\mu=\mathrm{M}$
$\therefore \mu=50$
Quartile deviation $=\frac{2}{3} \sigma$
$\therefore 15 x=\frac{2}{3} \sigma$
$ \therefore 15 \times \frac{3}{2}=\sigma$
$ \therefore \sigma=\frac{45}{2}=22.5$
The limits that include exactly middle $95.45 \%$ of the observations $= µ ± 2\sigma $
$= (µ – 2\sigma )$ to $(µ + 2\sigma )$
Putting, $µ = 50$ and $\sigma = 22.5$
The limits that include exactly middle $95.45 \%$ of the observations
$= [50 – 2(22.5)]$ to $(50 + 2(22.5)]$
$=(50 – 45)$ to $(50 + 45)$
$= 5$ to $95$
Hence, the limits that include exactly middle $95.45 \%$ of the observations for a normal distribution obtained is $(5, 95).$

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