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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.

Last updated date: 24th Jul 2024
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Hint: The probability of finding a student with a particular age depends on the frequency of students with the particular age. The number of total outcomes would be 15 (since, there are 15 students). Mean is given as \[\sum\limits_{i=1}^{n}{{{X}_{i}}}p({{X}_{i}})\], where $p({{X}_{i}})$ is the probability for occurrence of the event ${{X}_{i}}$. To calculate the variance, we define, E(X) = \[\sum\limits_{i=1}^{n}{{{X}_{i}}}p({{X}_{i}})\]. Thus, variance is $E({{X}^{2}})-{{(E(X))}^{2}}$. While, standard deviation is the square root of variance.

Complete step-by-step answer:
Now, first we start with finding the probability distribution of random variable X.

Probability = $\dfrac{\text{Desired number of outcomes}}{\text{Total number of outcomes}}$
Now, we can calculate probability for each event (namely the age of the students)-

P(X=14) = $\dfrac{2}{15}$, P(X=15) = $\dfrac{1}{15}$, P(X=16) = $\dfrac{2}{15}$, P(X=17) = $\dfrac{3}{15}=\dfrac{1}{5}$, P(X=18) = $\dfrac{1}{15}$, P(X=19) = $\dfrac{2}{15}$, P(X=20) = $\dfrac{3}{15}=\dfrac{1}{5}$, P(X=21) = $\dfrac{1}{15}$

This is the required probability distribution of the random variable X.

(These probabilities are calculated since we know that the total number of outcomes is 15 and the desired number of outcomes can be seen by the frequency of respective ages.)
Now, to find the mean,



=$14\times \dfrac{2}{15}+15\times \dfrac{1}{15}+16\times \dfrac{2}{15}+17\times \dfrac{3}{15}+18\times \dfrac{1}{15}+19\times \dfrac{2}{15}+20\times \dfrac{3}{15}+21\times \dfrac{1}{15}$


= 17.53 (approximately)

To find the variance,

We calculate $E({{X}^{2}})$

 E(${{X}^{2}}$) = \[\sum\limits_{i=1}^{n}{{{X}_{i}}^{2}}p({{X}_{i}})\]

E(${{X}^{2}}$) =\[{{14}^{2}}\times \dfrac{2}{15}+{{15}^{2}}\times \dfrac{1}{15}+{{16}^{2}}\times \dfrac{2}{15}+{{17}^{2}}\times \dfrac{3}{15}+{{18}^{2}}\times \dfrac{1}{15}+{{19}^{2}}\times \dfrac{2}{15}+{{20}^{2}}\times \dfrac{3}{15}+{{21}^{2}}\times \dfrac{1}{15}\]

E(${{X}^{2}}$) = $\dfrac{4683}{15}$

= 312.2

Now, the variance = $E({{X}^{2}})-{{(E(X))}^{2}}$

Variance = (312.2) -${{17.53}^{2}}$

(Since, mean = E(X) = 17.53)

Variance = 4.78 (approximately)

Now, standard deviation = \[\sqrt{\text{Variance}}\]=$\sqrt{4.78}$=2.186

Note: To calculate the probability of the event in case of discrete number of distribution, we simply calculate the desired frequency of the event and then divide it by the total number of outcomes. Variance can also be calculated by the formula $\dfrac{\sum\limits_{i=1}^{n}{{{({{X}_{i}}-\bar{X})}^{2}}}}{n}$. Here, n is the number of total outcomes and $\bar{X}$is the mean.