JEE MAIN - Mathematics (2020 - 6th September Evening Slot - No. 10)
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 _________ .
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
6
Explanation
Mean = $${{\sum {{x_1}.{f_1}} } \over {\sum {{f_1}} }}$$
= $${{0.{}^n{C_0} + 2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}} \over {{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}}$$
We know,
(1 + x)n = $${{}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ... + {}^n{C_n}{x^n}}$$ ...(1)
Put x = 2, at (1) we get
$$ \Rightarrow $$ 3n - 1 = $${2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}}$$
And Putting x = 1 in (1), we get
2n = $${{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}$$
$$ \therefore $$ Mean = $${{{3^n} - 1} \over {{2^n}}}$$
According to question,
$${{{3^n} - 1} \over {{2^n}}}$$ = $${{728} \over {{2^n}}}$$
$$ \Rightarrow $$ 3n = 729
$$ \Rightarrow $$ n = 6
= $${{0.{}^n{C_0} + 2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}} \over {{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}}$$
We know,
(1 + x)n = $${{}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ... + {}^n{C_n}{x^n}}$$ ...(1)
Put x = 2, at (1) we get
$$ \Rightarrow $$ 3n - 1 = $${2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}}$$
And Putting x = 1 in (1), we get
2n = $${{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}$$
$$ \therefore $$ Mean = $${{{3^n} - 1} \over {{2^n}}}$$
According to question,
$${{{3^n} - 1} \over {{2^n}}}$$ = $${{728} \over {{2^n}}}$$
$$ \Rightarrow $$ 3n = 729
$$ \Rightarrow $$ n = 6
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