Let the mean and variance of four numbers 3, 7, x and y (\(x>y\)) be 5 and 10 respectively. Then the mean of four numbers 3+2x, 7+2y, x+y and x-y is ___________.
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When solving for two variables from mean and variance, you will almost always get a system of equations involving \(x+y\) and \(x^2+y^2\). Use substitution to solve the system. Remember to check any additional conditions given, such as \(x>y\).
Step 1: Use Mean and Variance to Find x and y
Given numbers: 3, 7, x, y. Number of observations \(n=4\).
Mean \(\bar{X} = 5\).
\[ \bar{X} = \frac{3+7+x+y}{4} = 5 \implies 10+x+y = 20 \implies x+y=10 \]
Variance \(\sigma^2 = 10\). The formula for variance is \(\sigma^2 = \frac{\sum X_i^2}{n} - (\bar{X})^2\).
\[ 10 = \frac{3^2+7^2+x^2+y^2}{4} - (5)^2 \]
\[ 10 = \frac{9+49+x^2+y^2}{4} - 25 \]
\[ 35 = \frac{58+x^2+y^2}{4} \]
\[ 140 = 58+x^2+y^2 \implies x^2+y^2 = 82 \]
We have a system of two equations:
1) \(x+y=10 \implies y=10-x\)
2) \(x^2+y^2=82\)
Substitute (1) into (2):
\[ x^2 + (10-x)^2 = 82 \]
\[ x^2 + 100 - 20x + x^2 = 82 \]
\[ 2x^2 - 20x + 18 = 0 \]
\[ x^2 - 10x + 9 = 0 \]
\[ (x-9)(x-1) = 0 \]
So, \(x=9\) or \(x=1\).
Since \(x>y\):
If \(x=9\), then \(y=10-9=1\). This satisfies \(x>y\).
If \(x=1\), then \(y=10-1=9\). This does not satisfy \(x>y\).
So, we have \(x=9\) and \(y=1\). Step 2: Find the Mean of the New Set of Numbers
The new four numbers are: 3+2x, 7+2y, x+y, x-y.
Substitute \(x=9, y=1\):
The numbers are:
\(3+2(9) = 3+18 = 21\)
\(7+2(1) = 7+2 = 9\)
\(x+y = 9+1 = 10\)
\(x-y = 9-1 = 8\)
The new set of numbers is \{21, 9, 10, 8\}. Step 3: Calculate the New Mean
New Mean \(\bar{X}_{new} = \frac{21+9+10+8}{4} = \frac{48}{4} = 12\).
Alternatively,
New Mean = \(\frac{(3+2x)+(7+2y)+(x+y)+(x-y)}{4} = \frac{10+4x+2y}{4}\)
= \(\frac{10+4(9)+2(1)}{4} = \frac{10+36+2}{4} = \frac{48}{4}=12\). Step 4: Final Answer
The mean of the new four numbers is 12.
