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Home » Let X be a discrete random variable whose probability distribution is defined as follows: where k is a constant. Calculate (i) the value of k (ii) E (X) (iii) Standard deviation of X.
Let X be a discrete random variable whose probability distribution is defined as follows: where k is a constant. Calculate (i) the value of k (ii) E (X) (iii) Standard deviation of X.
CBSE | Class 12 | Exercise 1 | Maths | NCERT Exemplar | Probability
Let X be a discrete random variable whose probability distribution is defined as follows: where k is a constant. Calculate (i) the value of k (ii) E (X) (iii) Standard deviation of X.

(i) Given,

    \[P\left( X\text{ }=\text{ }x \right)\text{ }=\text{ }k\left( x\text{ }+\text{ }1 \right)\]

for

    \[x\text{ }=\text{ }1,\text{ }2,\text{ }3,\text{ }4\]

So,

    \[P\left( X\text{ }=\text{ }1 \right)\text{ }=\text{ }k\left( 1\text{ }+\text{ }1 \right)\text{ }=\text{ }2k\]

    \[P\left( X\text{ }=\text{ }2 \right)\text{ }=\text{ }k\left( 2\text{ }+\text{ }1 \right)\text{ }=\text{ }3k\]

    \[P\left( X\text{ }=\text{ }3 \right)\text{ }=\text{ }k\left( 3\text{ }+\text{ }1 \right)\text{ }=\text{ }4k\]

    \[P\left( X\text{ }=\text{ }4 \right)\text{ }=\text{ }k\left( 4\text{ }+\text{ }1 \right)\text{ }=\text{ }5k\]

Also,

    \[P\left( X\text{ }=\text{ }x \right)\text{ }=\text{ }2kx\]

for

    \[x\text{ }=\text{ }5,\text{ }6,\text{ }7\]

    \[P\left( X\text{ }=\text{ }5 \right)\text{ }=\text{ }2k\left( 5 \right)\text{ }=\text{ }10k\]

    \[P\left( X\text{ }=\text{ }6 \right)\text{ }=\text{ }2k\left( 6 \right)\text{ }=\text{ }12k\]

    \[P\left( X\text{ }=\text{ }7 \right)\text{ }=\text{ }2k\left( 7 \right)\text{ }=\text{ }14k\]

And, for otherwise it is

    \[0\]

.

Thus, the probability distribution is given by

We know that,

    \[\sum\limits_{i=1}^{n}{P({{X}_{i}})}=1\]

So,

    \[2k\text{ }+\text{ }3k\text{ }+\text{ }4k\text{ }+\text{ }5k\text{ }+\text{ }10k\text{ }+\text{ }12k\text{ }+\text{ }14k\text{ }=\text{ }1\]

⇒

    \[50k\text{ }=\text{ }1\]

    \[\Rightarrow k\text{ }=\text{ }1/50\]

Therefore, the value of k is

    \[1/50\]

.

(ii) Now, the probability distribution is

    \[E\left( X \right)\text{ }=\text{ }1\text{ }\times 2/50\text{ }+\text{ }2\text{ }\times 3/50\text{ }+\text{ }3\text{ }\times \text{ }4/50\text{ }+\text{ }4\text{ }\times 5/50\text{ }+\text{ }5\text{ }\times \text{ }10/50\text{ }+\text{ }6\text{ }\times \text{ }12/50\text{ }+\text{ }7\text{ }\times 14/50\]

=

    \[2/50\text{ }+\text{ }6/50\text{ }+\text{ }12/50\text{ }+\text{ }20/50\text{ }+\text{ }50/50\text{ }+\text{ }72/50\text{ }+\text{ }98/50\]

=

    \[260/50\]

=

    \[26/5\text{ }=\text{ }5.2\]

(ii) We know that the standard deviation (SD) = √Variance

Variance =

    \[E({{X}^{2}})\text{ }\text{ }{{\left[ E\left( X \right) \right]}^{2}}\]

i

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