The arithmetic mean of the data 0,1,2, , n with frequencies 1, ^ 4 C_ 1 ^ 4 C_ 2 , .., ^ ∞ C_ a is

Correct answer: D. n 2 Since, Mean = ∑ f_ i x_ i ∑ f_ i where x_ i are observations with frequencies f_ i , i=1,2, The required mean is given by aligned X &= 0.1+1. ^ n C_ 1 +2 ^ n C_ 2 + +n. ^ n C_ n 1+ ^ n C_ 1 + ^ n C_ 2 + + ^ n C_ n = ∑_ r=0 ^ n ^ n C_ r ∑_ r=0 ^ n ^ n C_ r = ∑_ r=1 ^ n r · n r ^ n-1 C_ r-1 ∑_ r=0 ^ n ^ n C_ r = n ∑ ∑ ^ n-1 C_ r-1 ∑_ r=0 ^ n ^ n C_ r = n.2^ n-1 2^ n = n 2 aligned

BITSAT 2015

Exam: BITSAT
Year: 2015
Subject: Mathematics
Chapter: Statistics

BITSAT 2015 · Mathematics · Statistics

The arithmetic mean of the data 0,1,2, , n with frequencies 1, ^ 4...

The arithmetic mean of the data

0,1,2, \ldots \ldots, n

with frequencies

1,{ }^{4} \mathrm{C}_{1}{ }^{4} \mathrm{C}_{2}, \ldots..,{ }^{\infty} \mathrm{C}_{\mathrm{a}}

Options

A $\mathrm{n}$
B $\frac{2^{n}}{n}$
C $\mathrm{n}+1$
D $\frac{\mathrm{n}}{2}$

Correct answer

D. $\frac{\mathrm{n}}{2}$

Step-by-step solution

Since, Mean

=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}

where

\mathrm{x}_{\mathrm{i}}

are observations with frequencies

\mathrm{f}_{\mathrm{i}}, \mathrm{i}=1,2,\ldots \ldots \ldots

The required mean is given by

\begin{aligned} \overline{\mathrm{X}} &=\frac{0.1+1.{ }^{\mathrm{n}} \mathrm{C}_{1}+2{ }^{\mathrm{n}} \mathrm{C}_{2}+\ldots \ldots+\mathrm{n}.{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}}{1+{ }^{\mathrm{n}} \mathrm{C}_{1}+{ }^{\mathrm{n}} \mathrm{C}_{2}+\ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}}\\ = \frac{\sum_{\mathrm{r}=0}^{\mathrm{n}} \mathrmr^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}=\frac{\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r} \cdot \frac{\mathrm{n}}{\mathrm{r}}{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}}{\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}\\ = \frac{\mathrm{n} \sum{\sum}^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}}{\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}=\frac{\mathrm{n}.2^{\mathrm{n}-1}}{2^{\mathrm{n}}}=\frac{\mathrm{n}}{2} \end{aligned}

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