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Step-by-Step Solution
Step 1: Express the given summation involving α
We are given:
\sum_{i=1}^{18} (X_i - \alpha) = 36.
This can be rewritten as:
\sum_{i=1}^{18} X_i - 18\alpha = 36.
Hence,
\sum_{i=1}^{18} X_i = 18\alpha + 36.
Let us denote
\sum_{i=1}^{18} X_i = 18(\alpha + 2).
Step 2: Express the given summation involving β
We are also given:
\sum_{i=1}^{18} (X_i - \beta)^2 = 90.
Expanding the square,
(X_i - \beta)^2 = X_i^2 - 2\beta X_i + \beta^2.
So the summation becomes:
\sum_{i=1}^{18} X_i^2 - 2\beta \sum_{i=1}^{18} X_i + 18\beta^2 = 90.
Using the earlier result
\sum_{i=1}^{18} X_i = 18(\alpha + 2),
we substitute:
\sum_{i=1}^{18} X_i^2 - 2\beta \bigl[18(\alpha + 2)\bigr] + 18\beta^2 = 90,
which implies:
\sum_{i=1}^{18} X_i^2 = 90 - 18\beta^2 + 36\beta(\alpha + 2).
Step 3: Use the standard deviation condition
The standard deviation of the 18 observations is given as 1. Recall that the variance σ^2 is:
\sigma^2 = \frac{1}{18} \sum_{i=1}^{18} \bigl(X_i - \overline{X}\bigr)^2,
where
\overline{X} = \frac{1}{18}\sum_{i=1}^{18} X_i.
Since \sigma^2 = 1 , we write:
1 = \frac{1}{18} \sum_{i=1}^{18} (X_i^2) - \Bigl(\overline{X}\Bigr)^2.
Substitute the expressions:
1) \sum_{i=1}^{18} X_i^2 = 90 - 18\beta^2 + 36\beta(\alpha + 2) ,
2) \overline{X} = \frac{1}{18} \cdot 18(\alpha + 2) = \alpha + 2.
Thus,
1 = \frac{1}{18} \bigl[\,90 - 18\beta^2 + 36\,\alpha\beta + 72\beta \bigr] - (\alpha + 2)^2.
Multiply the terms inside carefully:
1 = \frac{1}{18}(90) - \frac{1}{18}(18\beta^2) + \frac{1}{18}(36\,\alpha\beta) + \frac{1}{18}(72\beta)\;-\;(\alpha + 2)^2,
which simplifies to
1 = 5 - \beta^2 + 2\alpha\beta + 4\beta \;-\; (\alpha + 2)^2.
Step 4: Simplify the resulting expression
Next, expand (\alpha + 2)^2 = \alpha^2 + 4\alpha + 4 :
1 = 5 - \beta^2 + 2\alpha\beta + 4\beta - \alpha^2 - 4\alpha - 4.
Combine like terms:
1 = (5 - 4) - \alpha^2 - \beta^2 + 2\alpha\beta + 4\beta - 4\alpha.
So,
1 = 1 - \alpha^2 - \beta^2 + 2\alpha\beta + 4\beta - 4\alpha.
Cancel the 1’s on both sides:
0 = - \alpha^2 - \beta^2 + 2\alpha\beta + 4\beta - 4\alpha.
Or,
\alpha^2 - \beta^2 - 2\alpha\beta - 4\beta + 4\alpha = 0,
but more conveniently we can rearrange to spot factorization. A simpler rearrangement leads to:
\alpha^2 - \beta^2 + 2\alpha\beta + 4\beta - 4\alpha = 0.
Step 5: Factor the expression
Group terms involving (\alpha - \beta) :
(\alpha - \beta)(\alpha - \beta + 4) = 0.
Hence, the solutions for (\alpha - \beta) are:
\alpha - \beta = 0 \quad \text{or} \quad \alpha - \beta = -4.
But the problem states that \alpha and \beta are distinct, so
\alpha - \beta \neq 0.
Therefore,
\alpha - \beta = -4 \quad \Longrightarrow \quad |\alpha - \beta| = 4.
Final Answer
|\alpha - \beta| = 4.
