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.
Answer
363k+ views
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,
\[\sum\limits_{i=1}^{n}{{{X}_{i}}}p({{X}_{i}})\]
\[\sum\limits_{i=1}^{15}{{{X}_{i}}}p({{X}_{i}})\]
=$14\times \dfrac{2}{15}+15\times \dfrac{1}{15}+16\times \dfrac{2}{15}+17\times \dfrac{3}{15}+18\times \dfrac{1}{15}+19\times \dfrac{2}{15}+20\times \dfrac{3}{15}+21\times \dfrac{1}{15}$
=$\dfrac{1}{15}$(28+15+32+51+18+38+60+21)
=$\dfrac{263}{15}$
= 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.
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,
\[\sum\limits_{i=1}^{n}{{{X}_{i}}}p({{X}_{i}})\]
\[\sum\limits_{i=1}^{15}{{{X}_{i}}}p({{X}_{i}})\]
=$14\times \dfrac{2}{15}+15\times \dfrac{1}{15}+16\times \dfrac{2}{15}+17\times \dfrac{3}{15}+18\times \dfrac{1}{15}+19\times \dfrac{2}{15}+20\times \dfrac{3}{15}+21\times \dfrac{1}{15}$
=$\dfrac{1}{15}$(28+15+32+51+18+38+60+21)
=$\dfrac{263}{15}$
= 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.
Last updated date: 29th Sep 2023
•
Total views: 363k
•
Views today: 7.63k
Recently Updated Pages
What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

What is meant by shramdaan AVoluntary contribution class 11 social science CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is the value of 01+23+45+67++1617+1819+20 class 11 maths CBSE

How many millions make a billion class 6 maths CBSE

Draw a welllabelled diagram of a plant cell class 11 biology CBSE

Number of Prime between 1 to 100 is class 6 maths CBSE

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
