| Variate \[(x)\] | 0 | 1 | 2 | 3 | ? | N |
| Frequency \[(f)\] | \[^{n}{{C}_{0}}\] | \[^{n}{{C}_{1}}\] | \[^{n}{{C}_{2}}\] | \[^{n}{{C}_{3}}\] | ? | \[^{n}{{C}_{n}}\] |
A) \[\frac{(n+1)}{2}\]
B) \[\frac{n}{2}\]
C) \[\frac{{{2}^{n}}}{n}\]
D) None of the above
Correct Answer: B
Solution :
We have, \[\Sigma f{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{2}}+...{{+}^{n}}{{C}_{n}}={{2}^{n}}\] \[\Sigma f(x)={{0.}^{n}}{{C}_{0}}+{{1.}^{n}}{{C}_{1}}+{{2.}^{n}}{{C}_{2}}+{{3.}^{n}}{{C}_{3}}\] \[+.....+n{{.}^{n}}{{C}_{n}}\] \[=n+\frac{2n(n-1)}{2!}+\frac{3n(n-1)}{3!}+...+n.1\] \[=n\{1{{+}^{n-1}}{{C}_{1}}{{+}^{n-1}}{{C}_{2}}+...{{+}^{n-1}}{{C}_{n-1}}\}\] \[=n{{.2}^{n-1}}\] Thus, \[\overline{x}=\frac{\Sigma fx}{\Sigma f}=\frac{n{{.2}^{n-1}}}{{{2}^{n}}}=\frac{n}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
