Consider the data on x taking the values
0, 2, 4,
8,....., 2n with frequencies
nC0
,
nC1
,
nC2
,....,
nCn
respectively. If the
mean of this data is ${{728} \over {{2^n}}}$, then n is equal to _________ .
Answer (integer)
6
Solution
Mean = ${{\sum {{x_1}.{f_1}} } \over {\sum {{f_1}} }}$
<br><br>= $${{0.{}^n{C_0} + 2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}} \over {{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}}$$
<br><br>We know,
<br><br>(1 + x)<sup>n</sup> = ${{}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ... + {}^n{C_n}{x^n}}$ ...(1)
<br><br>Put x = 2, at (1) we get
<br>$\Rightarrow$ 3<sup>n</sup> - 1 = ${2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}}$
<br><br>And Putting x = 1 in (1), we get
<br><br>2<sup>n</sup> = ${{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}$
<br><br>$\therefore$ Mean = ${{{3^n} - 1} \over {{2^n}}}$
<br><br>According to question,
<br><br>${{{3^n} - 1} \over {{2^n}}}$ = ${{728} \over {{2^n}}}$
<br><br>$\Rightarrow$ 3<sup>n</sup> = 729
<br><br>$\Rightarrow$ n = 6
About this question
Subject: Mathematics · Chapter: Statistics · Topic: Measures of Central Tendency
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