Question

If the mean and variance of 5 values are both 4 and three out of 5 values are 1, 7 and 3, then the remaining two values are:

• A. 4 and 5
• B. 3 and 6
• C. 1 and 8
• D. 2 and 7

Hint:

Once you have the equation for the sum (\(x+y\)) and the sum of squares (\(x^2+y^2\)), you can quickly check the options. For option (A): \(x=4, y=5\). Sum = \(4+5=9\) (Correct). Sum of squares = \(4^2+5^2 = 16+25=41\) (Correct). This is a fast way to verify the answer in an exam.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are given the mean and variance of a dataset of 5 numbers, along with three of those numbers. We need to use the definitions of mean and variance to set up a system of two equations to solve for the two unknown numbers.

Step 2: Key Formula or Approach:
1. Mean: \(\mu = \frac{\sum x_i}{n}\) 2. Variance: \(\sigma^2 = \frac{\sum x_i^2}{n} - \mu^2\)

Step 3: Detailed Explanation:
Let the two unknown values be \(x\) and \(y\). We are given:

Number of values, \(n = 5\)
Mean, \(\mu = 4\)
Variance, \(\sigma^2 = 4\)
Known values are 1, 7, 3. \end{itemize} Using the Mean: \[ \mu = \frac{1 + 7 + 3 + x + y}{5} = 4 \] \[ 11 + x + y = 5 \times 4 \] \[ 11 + x + y = 20 \] \[ x + y = 9 \quad (Equation \ 1) \] Using the Variance: \[ \sigma^2 = \frac{1^2 + 7^2 + 3^2 + x^2 + y^2}{5} - \mu^2 = 4 \] \[ \frac{1 + 49 + 9 + x^2 + y^2}{5} - 4^2 = 4 \] \[ \frac{59 + x^2 + y^2}{5} - 16 = 4 \] \[ \frac{59 + x^2 + y^2}{5} = 20 \] \[ 59 + x^2 + y^2 = 100 \] \[ x^2 + y^2 = 41 \quad (Equation \ 2) \] Now we have a system of two equations with two variables: 1. \(x + y = 9\) 2. \(x^2 + y^2 = 41\) From Equation 1, we can write \(y = 9 - x\). Substitute this into Equation 2: \[ x^2 + (9-x)^2 = 41 \] \[ x^2 + (81 - 18x + x^2) = 41 \] \[ 2x^2 - 18x + 81 - 41 = 0 \] \[ 2x^2 - 18x + 40 = 0 \] Divide the equation by 2: \[ x^2 - 9x + 20 = 0 \] Factor the quadratic equation: \[ (x-4)(x-5) = 0 \] This gives two possible solutions for \(x\): \(x=4\) or \(x=5\). If \(x=4\), then \(y = 9 - 4 = 5\). If \(x=5\), then \(y = 9 - 5 = 4\). In either case, the remaining two values are 4 and 5.

Step 4: Final Answer:
The remaining two values are 4 and 5.