Question

The mean of 5 observations is 4.4 and their variance is 8.24. If three of those observations are 1, 2, and 6, then the other two observations are:

• A. 9, 4
• B. 9, 5
• C. 9, 2
• D. 9, 13

Hint:

To solve for unknown observations given the mean and variance, use the formulas for mean and variance, then solve the system of equations.

Answer 1

The Correct Option is A

We are given the following information: - The mean of 5 observations is 4.4 - The variance of 5 observations is 8.24 - Three of the observations are 1, 2, and 6. Let the five observations be \( x_1 = 1 \), \( x_2 = 2 \), \( x_3 = 6 \), and the unknown observations be \( x_4 \) and \( x_5 \). Step 1: Use the formula for the mean The formula for the mean of a set of observations is: \[ \frac{x_1 + x_2 + x_3 + x_4 + x_5}{5} = 4.4 \] Substitute the known values: \[ \frac{1 + 2 + 6 + x_4 + x_5}{5} = 4.4 \] \[ \frac{9 + x_4 + x_5}{5} = 4.4 \] Multiply both sides by 5: \[ 9 + x_4 + x_5 = 22 \] \[ x_4 + x_5 = 13 \] Step 2: Use the formula for the variance The formula for the variance is: \[ \text{Variance} = \frac{(x_1 - \mu)^2 + (x_2 - \mu)^2 + (x_3 - \mu)^2 + (x_4 - \mu)^2 + (x_5 - \mu)^2}{5} \] Where \( \mu \) is the mean. We are given the variance as 8.24, and \( \mu = 4.4 \). Substituting the values, we get: \[ \frac{(1 - 4.4)^2 + (2 - 4.4)^2 + (6 - 4.4)^2 + (x_4 - 4.4)^2 + (x_5 - 4.4)^2}{5} = 8.24 \] Simplify the terms: \[ (1 - 4.4)^2 = 11.56, \quad (2 - 4.4)^2 = 5.76, \quad (6 - 4.4)^2 = 2.56 \] Substitute these into the variance equation: \[ \frac{11.56 + 5.76 + 2.56 + (x_4 - 4.4)^2 + (x_5 - 4.4)^2}{5} = 8.24 \] \[ \frac{19.88 + (x_4 - 4.4)^2 + (x_5 - 4.4)^2}{5} = 8.24 \] Multiply both sides by 5: \[ 19.88 + (x_4 - 4.4)^2 + (x_5 - 4.4)^2 = 41.2 \] \[ (x_4 - 4.4)^2 + (x_5 - 4.4)^2 = 21.32 \] Step 3: Solve for \( x_4 \) and \( x_5 \) We already know that \( x_4 + x_5 = 13 \). Let’s check for values of \( x_4 \) and \( x_5 \) that satisfy both \( x_4 + x_5 = 13 \) and the variance equation. After testing possible pairs, we find that the values \( x_4 = 9 \) and \( x_5 = 4 \) satisfy both conditions. Thus, the other two observations are \( x_4 = 9 \) and \( x_5 = 4 \). The correct answer is option (1), 9, 4.

Answer 2

Given:
- Mean of 5 observations = 4.4
- Variance of 5 observations = 8.24
- Three observations are: 1, 2, and 6

Step 1: Let the other two observations be x and y
So, the five observations are: 1, 2, 6, x, y

Step 2: Use the formula for mean
Mean = (Sum of observations) / 5 = 4.4
So, total sum = 4.4 × 5 = 22
Now, sum of known observations = 1 + 2 + 6 = 9
Therefore, x + y = 22 - 9 = 13 ... (Equation 1)

Step 3: Use the formula for variance
Variance = (Sum of squares of observations) / 5 - (Mean)²
Given variance = 8.24 and mean = 4.4
So,
8.24 = [ (1² + 2² + 6² + x² + y²) / 5 ] - (4.4)²
→ 8.24 = [ (1 + 4 + 36 + x² + y²) / 5 ] - 19.36
→ 8.24 + 19.36 = (41 + x² + y²) / 5
→ 27.6 × 5 = 41 + x² + y²
→ 138 = 41 + x² + y²
→ x² + y² = 97 ... (Equation 2)

Step 4: Solve the system of equations
From Equation 1: x + y = 13 → y = 13 - x
Substitute into Equation 2:
x² + (13 - x)² = 97
x² + (169 - 26x + x²) = 97
2x² - 26x + 169 = 97
2x² - 26x + 72 = 0
Divide entire equation by 2:
x² - 13x + 36 = 0
Solve the quadratic:
x = [13 ± √(169 - 144)] / 2 = [13 ± √25] / 2 = [13 ± 5] / 2
So, x = 9 or 4
Then, y = 4 or 9

Final Answer:
The other two observations are: 9 and 4