question_answer

A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 yr. One student is selected in such a manner that each has the same chance of being of 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.

OR

3 bad eggs are mixed with 7 good eggs 3 eggs are taken at random from the 10 eggs. Find the probability distribution of number of bad eggs drawn. Also, find the mean and variance of the probability distribution.

Answer:

Here, total number of students = 15

The ages of students in ascending order are 14, 14, 15, 16, 16, 17, 17, 17, 18, 19, 19, 20, 20, 20, 21

Now,     \[P\,(X=14)=\frac{2}{15},\,\,P\,(X=15)=\frac{1}{15},\]

\[P\,(X=16)=\frac{2}{15},\,\,P\,(X=17)=\frac{3}{15}\]

\[P\,(X=18)=\frac{1}{15},\,\,P\,(X=19)=\frac{2}{15}\]

\[P\,(X=20)=\frac{3}{15},\,\,P\,(X=21)=\frac{1}{15}\]

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

X 14 15 16 17 18 19 20 21 Number of students 2 1 2 3 1 2 3 1 \[P\,(X)\] \[\frac{2}{15}\] \[\frac{1}{15}\] \[\frac{2}{15}\] \[\frac{3}{15}\] \[\frac{1}{15}\] \[\frac{2}{15}\] \[\frac{3}{15}\] \[\frac{1}{15}\]

The third row gives the probability distribution of X.

Mean\[X=\sum X\,\,P\,(X)\]

\[=\frac{\left[ \begin{align}   & 14\times 2+15\times 1+16\times 2+17\times 3+18\times 1+19\times 2 \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+20\times 3+21\times 1 \\ \end{align} \right]}{15}\]

\[=\frac{28+15+32+51+18+38+60+21}{15}=\frac{263}{15}=17.53\]Variance\[X=\sum {{X}^{2}}P\,(X)-{{(Mean)}^{2}}\]

\[=\frac{\left[ \begin{align}   & {{(14)}^{2}}\times 2+{{(15)}^{2}}\times 1+{{(16)}^{2}}\times 2+{{(17)}^{2}}\times 3 \\  & \,\,\,\,\,+{{(18)}^{2}}\times 1+{{(19)}^{2}}+{{(20)}^{2}}\times 3+{{(21)}^{2}}\times 1 \\ \end{align} \right]}{15}\]

\[-{{\left( \frac{263}{15} \right)}^{2}}\]

\[=\frac{392+225+512+867+324+700+1200+441}{15}\]\[-{{\left( \frac{263}{15} \right)}^{2}}\]

\[=\frac{4683}{15}-{{\left( \frac{263}{15} \right)}^{2}}\]

\[=312.2-307.4=4.8\]

SD of\[X=\sqrt{\text{Variance}}=\sqrt{4.8}=2.19\]

OR

Let X denotes the random variable showing the number of bad eggs.

Then, X can take the value 0, 1, 2 or 3.

\[\therefore \]\[P\,(X=0)=P\](none of the egg is defective)

= P (all the three eggs are good ones)

\[=\frac{{}^{7}{{C}_{3}}}{{}^{10}{{C}_{3}}}=\left( \frac{7\times 6\times 5}{3\times 2\times 1}\times \frac{3\times 2\times 1}{10\times 9\times 8} \right)=\frac{7}{24}\]

\[P\,(X=1)=P\](1 bad and 2 good eggs)

\[=\frac{{}^{3}{{C}_{1}}\times {}^{7}{{C}_{2}}}{{}^{10}{{C}_{3}}}=3\times \frac{7\times 6}{2\times 1}\times \frac{3\times 2\times 1}{10\times 9\times 8}=\frac{21}{40}\]

\[P\,(X=2)=P\](2 bad and 1 good egg)

\[=\frac{{}^{3}{{C}_{2}}\times {}^{7}{{C}_{1}}}{{}^{10}{{C}_{3}}}=\frac{3\times 2}{2\times 1}\times 7\times \frac{3\times 2\times 1}{10\times 9\times 8}=\frac{7}{40}\]

\[P\,(X=3)=P\](3 bad eggs)

\[=\frac{{}^{3}{{C}_{3}}}{{}^{10}{{C}_{3}}}=1\times \frac{3\times 2\times 1}{10\times 9\times 8}\]

\[=\frac{1}{20}\]

Thus, the probability distribution is given by

\[X={{x}_{i}}\] 0 1 2 3 \[{{p}_{i}}\] \[\frac{7}{24}\] \[\frac{21}{40}\] \[\frac{7}{40}\] \[\frac{1}{120}\]

\[\therefore \]Mean \[(\mu )=\sum {{x}_{i}}{{p}_{i}}=0\times \frac{7}{24}+1\times \frac{21}{40}\]

\[+2\times \frac{7}{40}+3\times \frac{1}{120}=\frac{9}{10}=0.9\]

Variance,\[{{\sigma }^{2}}=\sum x_{i}^{2}{{p}_{i}}-{{\mu }^{2}}\]

\[=\left( 0\times \frac{7}{24} \right)+\left( 1\times \frac{21}{40} \right)+\left( 4\times \frac{7}{40} \right)\]

\[+\left( 9\times \frac{1}{120} \right)-\frac{81}{100}\]

\[=\frac{13}{10}-\frac{81}{100}\]

\[=\frac{49}{100}=0.49\]