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Home  >>  CBSE XII  >>  Math  >>  Probability
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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.

$\begin{array}{1 1}Mean = 17.53, Variance = 4.89, Standard Deviation = 2.21 \\Mean = 18.53, Variance = 4.89, Standard Deviation = 2.14 \\Mean = 17.53, Variance = 5.89, Standard Deviation = 2.21 \\ Mean = 17.53, Variance = 5.89, Standard Deviation = 2.14 \end{array} $

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Toolbox:
  • Mean of the probability distribution = $\sum (X_i \times P(X_i))$ The Expected value of X is nothing but the mean of X.
  • Standard Deviation = $\sqrt{\text{Variance}}$, where Variance $= E (X^2) - E(X)^2$
There are 15 students whose ages are: 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. The frequency and probability distribution of this set can be represented as follows:
\begin{Bmatrix} \text{X} & 14 &15 & 16 & 17 &18 & 19 & 20&21 \\ \text{Frequency} & 2 & 1& 2 & 3& 1 & 2& 3& 1\\ \text{P(X)} & \large\frac{2}{15}& \large\frac{1}{15}&\large\frac{2}{15}&\large\frac{5}{15}&\large\frac{1}{15}&\large\frac{2}{15}&\large\frac{3}{15}&\large\frac{1}{15} \end{Bmatrix}
Mean of the probability distribution = $\sum (X_i \times P(X_i))$
Mean or $E(X) = \sum (X_i \times P(X_i)) = $$14 \times \large\frac{2}{15} + $$ 15\times\large\frac{1}{15} + $$16\times\large\frac{2}{15} + $$17\times\large\frac{3}{15}+ $$18\times\large\frac{1}{15} + $$19\times \large\frac{2}{15} + $$ 20\times\large\frac{3}{15} + $$21\times\large\frac{1}{15} $
$E (X) = \large \frac{28+15+32+51+18+28+60+21}{15} = \frac{263}{15} =$$ 17.53$
$E(X^2) = \sum ((X_i)^2\times P(X_i)) = $$14^2 \times \large\frac{2}{15} + $$ 15^2\times\large\frac{1}{15} + $$16^2\times\large\frac{2}{15} + $$17^2\times\large\frac{3}{15}+ $$18^2\times\large\frac{1}{15} + $$19^2\times \large\frac{2}{15} + $$ 20^2\times\large\frac{3}{15} + $$21^2\times\large\frac{1}{15} $
$E (X^2) = \large \frac{392+225+512+876+324+722+1200+441}{15} = \frac{4683}{15} = $$312.2$
Therefore Variance = $312.2 - 17.53^2 = 4.89$
Therefore Standard Deviation = $\sqrt{4,89} = 2.21$
answered Jun 20, 2013 by balaji.thirumalai
 

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