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RANDOM VARIABLE ANE DISCRETE PROBABILITY DISTRIBUTION — Statistics STD 12 Commerce — Question

Gujarat BoardEnglish MediumSTD 12 CommerceStatisticsRANDOM VARIABLE ANE DISCRETE PROBABILITY DISTRIBUTION3 Marks
Question
If the following distribution is a probability distribution of variable $X$, then find constant $k . P(x)=\frac{6-|x-7|}{k} ; x=4,5,6,7,8,9,10$
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Answer

Here, $\mathrm{P}(\mathrm{x})=\frac{6-|x-7|}{k}$
Putting, $x=4,5,6,7,8,9,10$
$\mathrm{P}(4)=\frac{6-|4-7|}{k}=\frac{3}{k}$
$ \mathrm{P}(5)=\frac{6-|5-7|}{k}=\frac{4}{k}$
$ \mathrm{P}(6)=\frac{6-|6-7|}{k}=\frac{5}{k}$
$ \mathrm{P}(7)=\frac{6-|7-7|}{k}=\frac{6}{k}$
$ \mathrm{P}(8)=\frac{6-|8-7|}{k}=\frac{5}{k}$
$ \mathrm{P}(9)=\frac{6-|9-7|}{k}=\frac{4}{k}$
$ \mathrm{P}(10)=\frac{6-|10-7|}{k}=\frac{3}{k}$
Now, by condition of probability distribution,
we must have
$\sum \mathrm{p}(\mathrm{x})=1$
$ \therefore \mathrm{P}(4)+\mathrm{p}(5)+\mathrm{P}(6)+\mathrm{P}(7)+\mathrm{P}(8)+\mathrm{p}(9)+\mathrm{P}(10)=1$
$ \therefore \frac{3}{k}+\frac{4}{k}+\frac{5}{k}+\frac{6}{k}+\frac{5}{k}+\frac{4}{k}+\frac{3}{k}$
$ \therefore \frac{30}{k}=1$
$ \therefore \mathrm{k}=30$
Hence, the value of $k$ obtained is $30 .$

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