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If the mean of a set of observations \mathrm{d_{1},d_{2},.....d_{n}} is \mathrm{d}, then the mean of the observations \mathrm{d_{i+2^{i}}} where \mathrm{i= 1,2,...,n,} is:

Option: 1

\mathrm{d+i+2^i}


Option: 2

\mathrm{d+2^i}


Option: 3

\mathrm{d \overline{+}\left(2^i+1\right)}


Option: 4

\mathrm{d+2^i}


Answers (1)

best_answer

The mean of the observations \mathrm{d_{\left(i+2^i\right)}, where \, i=1,2, \ldots, n}, can be calculated as follows:
First, let's find the number of terms in the new set of observations. For each i, we have one term, so the number of terms is \mathrm{ n}.

Next, we calculate the sum of the new observations:

\mathrm{ \text { Sum }=d_{\left(1+2^1\right)}+d_{\left(2+2^2\right)}+\ldots+d_{\left(n+2^n\right)}}
To simplify this expression, we can rewrite it as:

\mathrm{ \text { Sum }=d_{(3)}+d_{(6)}+\ldots+d_{\left(n+2^n\right)}}

Now, let's express the indices in terms of i:
For \mathrm{ i=1},the index is 3 .
For \mathrm{ i=2}, the index is 6 .
For \mathrm{ i=3}, the index is 11 .

And so on.
We can observe that the indices follow a pattern given by the expression \mathrm{ 2+2^{(i+1)}}.

Therefore, the sum of the new observations can be written as:
\mathrm{ \text { Sum }=d_{\left(2+2^{(1+1)}\right)}+d_{\left(2+2^{(2+1)}\right)}+\ldots+d_{\left(2+2^{(n+1)}\right)}}

Simplifying further, we have:
\mathrm{ \text { Sum }=d_{\left(2+2^2\right)}+d_{\left(2+2^3\right)}+\ldots+d_{\left(2+2^{(n+1)}\right)}}

Now, let's consider the mean of the new observations:
\mathrm{ \begin{aligned} & \text { Mean }=\text { Sum } / n \\ & \text { Mean }=\left(d_{\left(2+2^2\right)}+d_{\left(2+2^3\right)}+\ldots+d_{\left(2+2^{(n+1)}\right)}\right) / n \end{aligned}}
Therefore, the correct answer is \mathrm{ d+2^{i}}.

Posted by

Ajit Kumar Dubey

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