Question

The mean and variance of the data \(4, 5, 6, 6, 7, 8, x, y\), where \(x<y\), are \(6\) and \(\frac{9}{4}\), respectively. Then \(x^4+y^2\) is equal to

• A. 162
• B. 320
• C. 674
• D. 420

Answer 1

The Correct Option is B

To solve this problem, we need to determine \(x^4 + y^2\) given that the mean and variance of the data set \(4, 5, 6, 6, 7, 8, x, y\) are 6 and \(\frac{9}{4}\), respectively, where \(x < y\).  

1. First, calculate the mean: The mean is given by the formula: \(Mean = \frac{4 + 5 + 6 + 6 + 7 + 8 + x + y}{8} = 6\) 
   Simplifying this equation, we get: \(4 + 5 + 6 + 6 + 7 + 8 + x + y = 48\) 
   \(36 + x + y = 48\) 
   \(x + y = 12\)
2. Next, consider the variance formula: \(Variance = \frac{\sum{(x_i - \bar{x})^2}}{n}\) 
   Where \(\bar{x} = 6\), \(n = 8\), and given variance is \(\frac{9}{4}\): \(\frac{(4 - 6)^2 + (5 - 6)^2 + (6 - 6)^2 + (6 - 6)^2 + (7 - 6)^2 + (8 - 6)^2 + (x - 6)^2 + (y - 6)^2}{8} = \frac{9}{4}\) 
   Simplifying: \(\frac{4 + 1 + 0 + 0 + 1 + 4 + (x - 6)^2 + (y - 6)^2}{8} = \frac{9}{4}\) 
   \(10 + (x - 6)^2 + (y - 6)^2 = 18\) 
   \((x - 6)^2 + (y - 6)^2 = 8\)
3. Now, we solve the equations:
   • From \(x + y = 12\), express \(y\) in terms of \(x\): \(y = 12 - x\)
   • Substitute \(y = 12 - x\) into \((x-6)^2 + (y-6)^2 = 8\): \((x-6)^2 + (12-x-6)^2 = 8\) 
     \((x-6)^2 + (6-x)^2 = 8\) 
     Without loss of generality, expand and simplify: \((x-6)^2 + (6-x)^2 = 8\) results in a quadratic equation: \(2(x-6)^2 = 8\) 
     x - 6 = 2 or \(x - 6 = -2\) 
     If \(x - 6 = 2\), then \(x = 8\) 
     If \(x - 6 = -2\), then \(x = 4\)
4. Substituting into \(x + y = 12\):
   • If \(x = 8\), \(y = 4\) (which is invalid since \(x < y\)).
   • If \(x = 4\), then \(y = 8\)
5. Thus, given \(x = 4\), \(y = 8\) satisfies the condition \(x < y\). 
   Compute \(x^4 + y^2\): \(x^4 + y^2 = 4^4 + 8^2 = 256 + 64 = 320\)

The correct answer is 320.