18. It is given that the six numbers, $$ 10,6,18, x, 15 $$ and $$ y $$ have a mean of 9 and a standard deviation of 6 . Find the value of $$ x $$ and of $$ y $$.

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We are given the six numbers
$$
10,\; 6,\; 18,\; x,\; 15,\; y
$$
with a mean of $$9$$ and a standard deviation of $$6$$.

**Step 1. Find the sum of all numbers using the mean.**

Since the mean is $$9$$, the sum of the six numbers is
$$
6 \times 9 = 54.
$$
Thus,
$$
10 + 6 + 18 + x + 15 + y = 54.
$$
Adding the known numbers,
$$
10 + 6 + 18 + 15 = 49,
$$
we have
$$
49 + x + y = 54 \quad \Longrightarrow \quad x + y = 5.
$$

**Step 2. Relate the standard deviation to the sum of squared deviations.**

The standard deviation is given as $$6$$. Assuming this is the population standard deviation, the variance is
$$
6^2 = 36.
$$
The variance for a population is computed as the average of squared deviations from the mean, so:
$$
\frac{1}{6} \sum (a_i - 9)^2 = 36.
$$
Thus, the total sum of squared deviations is
$$
\sum (a_i - 9)^2 = 6 \times 36 = 216.
$$

**Step 3. Compute the sum of squared deviations for the known numbers.**

For each known number, compute $$(\text{number} - 9)^2$$:
- For $$10$$: $$10 - 9 = 1$$ $$\Rightarrow$$ $$1^2 = 1$$,
- For $$6$$: $$6 - 9 = -3$$ $$\Rightarrow$$ $$(-3)^2 = 9$$,
- For $$18$$: $$18 - 9 = 9$$ $$\Rightarrow$$ $$9^2 = 81$$,
- For $$15$$: $$15 - 9 = 6$$ $$\Rightarrow$$ $$6^2 = 36$$.

Adding these together,
$$
1 + 9 + 81 + 36 = 127.
$$

**Step 4. Find the sum of squared deviations for $$x$$ and $$y$$.**

Let the squared deviations for $$x$$ and $$y$$ be:
$$
(x-9)^2 + (y-9)^2.
$$
Since the total must be $$216$$,
$$
(x-9)^2 + (y-9)^2 = 216 - 127 = 89.
$$

**Step 5. Substitute $$y$$ in terms of $$x$$ into the squared deviations equation.**

We already have
$$
x + y = 5 \quad \Longrightarrow \quad y = 5 - x.
$$
Substitute into the equation:
$$
(x-9)^2 + ((5-x)-9)^2 = 89.
$$
Simplify the second term:
$$
(5 - x - 9) = -x - 4,
$$
so the equation becomes:
$$
(x-9)^2 + (-x-4)^2 = 89.
$$
Note that $$(-x-4)^2 = (x+4)^2$$. Thus,
$$
(x-9)^2 + (x+4)^2 = 89.
$$

**Step 6. Expand and simplify the equation.**

Expanding each term:
$$
(x-9)^2 = x^2 - 18x + 81,
$$
$$
(x+4)^2 = x^2 + 8x + 16.
$$
Adding these,
$$
x^2 - 18x + 81 + x^2 + 8x + 16 = 2x^2 - 10x + 97.
$$
Setting equal to $$89$$:
$$
2x^2 - 10x + 97 = 89.
$$
Subtract $$89$$ from both sides:
$$
2x^2 - 10x + 8 = 0.
$$
Divide the entire equation by $$2$$:
$$
x^2 - 5x + 4 = 0.
$$

**Step 7. Solve the quadratic equation.**

The quadratic factors as:
$$
(x-1)(x-4)=0.
$$
Thus, the solutions are:
$$
x = 1 \quad \text{or} \quad x = 4.
$$

Using $$x + y = 5$$, if:
- $$x = 1$$ then $$y = 5 - 1 = 4$$,
- $$x = 4$$ then $$y = 5 - 4 = 1$$.

**Final Answer:**

The numbers $$x$$ and $$y$$ are $$1$$ and $$4$$ (in either order).
$$ x = 1 $$ and $$ y = 4 $$

It is given that the six numbers, and
have a mean of 9 and a standard deviation of 6 .
Find the value of and of .

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It is given that the six numbers, and have a mean of 9 and a standard deviation of 6 . Find the value of and of .