Probability Ncert Solutions Maths class 12th Maths Class 12th Probability

63. A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.

6 Views | Posted 3 months ago

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

4 months ago

63. There are 15 students in the class. Each student has the same chance to be chosen. Therefore, the probability of each student to be selected is 1/15.

The given information can be compiled in the frequency table as follows.

14	15	16	17	18	19	20	21
2	1	2	3	1	2	3	1

P(X = 14) =2/15, P(X = 15) =1/15, P(X = 16) =2/15, P(X = 16) =3/15,

P(X = 18) =115, P(X = 19) =2/15, P(X = 20) =3/15, P(X = 21) =1/15

Therefore, the probability distribution of random variable X is as follows.

X	14	15	16	17	18	19	20	21
f	2/15	1/15	2/15	3/15	1/15	2/15	3/15	1/15

Then, mean of X = E(X)

$\begin{array}{l} = \sum_{}^{} X_{i} P (X_{i}) \\ = 14 \times \frac{2}{15} + 15 \times \frac{1}{15} + 16 \times \frac{2}{15} + 17 \times \frac{3}{15} + 18 \times \frac{1}{15} + 19 \times \frac{2}{15} + 20 \times \frac{3}{15} + 21 \times \frac{1}{15} \\ = \frac{1}{15} (28 + 15 + 32 + 51 + 18 + 38 + 60 + 21) \\ = \frac{263}{15} \\ = 17.53 \end{array}$

$\begin{array}{l} E (X^{2}) = \sum_{}^{} {X^{2}}_{i} P (X_{i}) \\ = {(14)}^{2} . \frac{2}{15} + {(15)}^{2} . \frac{1}{15} + {(16)}^{2} . \frac{2}{15} + {(17)}^{2} . \frac{3}{15} + {(18)}^{2} . \frac{1}{15} + {(19)}^{2} . \frac{2}{15} + {(20)}^{2} . \frac{3}{15} + {(21)}^{2} . \frac{1}{15} \\ = \frac{1}{15} . (392 + 225 + 512 + 867 + 324 + 722 + 1200 + 441) \\ = \frac{4683}{15} \\ = 312.2 \end{array}$

$\begin{array}{l} ∴ V a r i a n c e (X) = E (X^{2}) - {[E (X)]}^{2} \\ = 312.2 - {(\frac{263}{15})}^{2} \\ = 312.2 - 307.4177 \\ = 4.7823 \\ \approx 4.78 \end{array}$

Standard derivation $\begin{array}{l} =\sqrtvariance (X) \\ =\sqrt4.78 \\ = 2.186 \approx 2.19 \end{array}$

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