Question

If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then \(a + b + ab\) is equal to:

• A. 105
• B. 103
• C. 100
• D. 106

Hint:

In solving problems involving variance and mean, use the formula for variance to derive the necessary equations. Then, use the system of equations to solve for unknowns such as \( a + b + ab \).

Answer 1

The Correct Option is B

This problem provides a set of 8 observations, including two unknown values, \(a\) and \(b\). We are given the mean and variance of this dataset and are asked to calculate the value of the expression \( a + b + ab \).

Concept Used:

The solution is based on the definitions of the mean and variance of a dataset.

1. Mean (\( \bar{x} \)): The mean of a set of \( n \) observations \( x_1, x_2, \ldots, x_n \) is the sum of the observations divided by the number of observations. \[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \]
2. Variance (\( \sigma^2 \)): The variance measures the spread of the data. It is calculated as the mean of the squares of the observations minus the square of the mean. \[ \sigma^2 = \frac{\sum_{i=1}^{n} x_i^2}{n} - (\bar{x})^2 \]

We will use these formulas to set up two equations with two unknowns, \(a\) and \(b\), and solve for the quantities needed.

Step-by-Step Solution:

Step 1: Use the formula for the mean to find the sum \( a + b \).

The given observations are {6, 4, a, 8, b, 12, 10, 13}.

The number of observations is \( n = 8 \).

The given mean is \( \bar{x} = 9 \).

The sum of all observations is:

\[ \sum x_i = 6 + 4 + a + 8 + b + 12 + 10 + 13 = 53 + a + b \]

Using the mean formula, \( \bar{x} = \frac{\sum x_i}{n} \):

\[ 9 = \frac{53 + a + b}{8} \]

Multiplying both sides by 8:

\[ 72 = 53 + a + b \]

Solving for \( a + b \):

\[ a + b = 72 - 53 = 19 \]

Step 2: Use the formula for the variance to find the sum of squares \( a^2 + b^2 \).

The given variance is \( \sigma^2 = 9.25 \).

The sum of the squares of the observations is:

\[ \sum x_i^2 = 6^2 + 4^2 + a^2 + 8^2 + b^2 + 12^2 + 10^2 + 13^2 \] \[ \sum x_i^2 = 36 + 16 + a^2 + 64 + b^2 + 144 + 100 + 169 = 529 + a^2 + b^2 \]

Using the variance formula, \( \sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 \):

\[ 9.25 = \frac{529 + a^2 + b^2}{8} - (9)^2 \] \[ 9.25 = \frac{529 + a^2 + b^2}{8} - 81 \]

Add 81 to both sides:

\[ 90.25 = \frac{529 + a^2 + b^2}{8} \]

Multiply both sides by 8:

\[ 90.25 \times 8 = 529 + a^2 + b^2 \] \[ 722 = 529 + a^2 + b^2 \]

Solving for \( a^2 + b^2 \):

\[ a^2 + b^2 = 722 - 529 = 193 \]

Step 3: Use the values of \( a + b \) and \( a^2 + b^2 \) to find the product \( ab \).

We use the algebraic identity \( (a+b)^2 = a^2 + b^2 + 2ab \).

Substitute the values we have found:

\[ (19)^2 = 193 + 2ab \] \[ 361 = 193 + 2ab \] \[ 2ab = 361 - 193 = 168 \] \[ ab = \frac{168}{2} = 84 \]

Final Computation & Result

The problem asks for the value of the expression \( a + b + ab \).

Using the results from our calculations:

\[ a + b = 19 \] \[ ab = 84 \]

Therefore:

\[ a + b + ab = 19 + 84 = 103 \]

The value of \( a + b + ab \) is 103.