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Let `barx` be the mean of x_{1}, x_{2}, ..., x_{n} and y the mean of y_{1}, y_{2}, ..., y_{n}. If z is the mean of x_{1}, x_{2}, ..., x_{n}, y_{1}, y_{2}, ..., y_{n}, then z is equal to ______.

#### Options

`barx + bary`

`(barx + bary)/2`

`(barx + bary)/n`

`(barx + bary)/(2n)`

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#### Solution

Let `barx` be the mean of x_{1}, x_{2}, ..., x_{n} and y the mean of y_{1}, y_{2}, ..., y_{n}. If z is the mean of x_{1}, x_{2}, ..., x_{n}, y_{1}, y_{2}, ..., y_{n}, then z is equal to **`(barx + bary)/2`**.

**Explanation:**

Given, `sum_(i = 1)^n x_i = nbarx` and `sum_(i = 1)^n y_i = nbary` .....(i) `[because barx = (sum_(i = 1)^n x_i)/n]`

Now, `barz = ((x_1 + x_2 + ... + x_n) + (y_1 + y_2 + ... + y_n))/(n + n)`

= `(sum_(i = 1)^n x_i + sum_(i = 1)^n y_i)/(2n)`

= `(nbarx + nbary)/(2n)`

= `(barx + bary)/2` ......[From equation (i)]

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