Statistics And Probability Question 301
Question: The mean of the values 0, 1, 2,……,n having corresponding weight $ ^{n}C_0,{{,}^{n}}C_1,{{,}^{n}}C_2,……..,,{{,}^{n}}C_{n} $ respectively is
[AMU 1990; CET 1998]
Options:
A) $ \frac{2^{n}}{n+1} $
B) $ \frac{{2^{n+1}}}{n(n+1)} $
C) $ \frac{n+1}{2} $
D) $ \frac{n}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
The required mean is $ \bar{x}=\frac{0.1+{{1.}^{n}}C_1+{{2.}^{n}}C_2+{{3.}^{n}}C_3+……+n{{.}^{n}}C_{n}}{1{{+}^{n}}C_1{{+}^{n}}C_2+….{{+}^{n}}C_{n}} $ $ =\frac{\sum\limits_{r=0}^{n}{r.,{{,}^{n}}C_{r}}}{\sum\limits_{r=0}^{n}{^{n}C_{r}}}=\frac{\sum\limits_{r=1}^{n}{r.\frac{n}{r},{{,}^{n-1}}{C_{r-1}}}}{\sum\limits_{r=0}^{n}{^{n}C_{r}}} $ = $ \frac{n\sum\limits_{r=1}^{n}{^{n-1}{C_{r-1}}}}{\sum\limits_{r=0}^{n}{^{n}C_{r}}} $ $ =\frac{n{{.2}^{n-1}}}{2^{n}} $ $ =\frac{n}{2} $ .
BETA
