The problem asks us to find the value of the expression $a+b+ab$ given a dataset and its mean and variance.

Dataset and Given Information

The dataset is {6, 4, a, 8, b, 12, 10, 13}.

• Number of observations, $n = 8$.
• Given Mean, $\bar{x} = 9$.
• Given Variance, $\sigma^2 = 9.25$.

Calculating Sum of Observations

First, let's find the sum of the known numbers in the dataset: $6 + 4 + 8 + 12 + 10 + 13 = 53$.

The sum of all observations is $\sum x_i = 53 + a + b$.

Using the Mean Formula

The formula for the mean is $\bar{x} = \frac{\sum x_i}{n}$.

Substituting the given values:

$9 = \frac{53 + a + b}{8}$

Multiply both sides by 8:

$9 \times 8 = 53 + a + b$

$72 = 53 + a + b$

Rearrange to find the sum $a+b$:

$a + b = 72 - 53$

$a + b = 19$ (Equation 1)

Calculating Sum of Squares

Next, let's find the sum of the squares of the known numbers:

$6^2 + 4^2 + 8^2 + 12^2 + 10^2 + 13^2 = 36 + 16 + 64 + 144 + 100 + 169 = 529$.

The sum of the squares of all observations is $\sum x_i^2 = 529 + a^2 + b^2$.

Using the Variance Formula

The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.

Substitute the known values:

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

$9.25 = \frac{529 + a^2 + b^2}{8} - 81$

Add 81 to both sides:

$9.25 + 81 = \frac{529 + a^2 + b^2}{8}$

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

Multiply both sides by 8:

$90.25 \times 8 = 529 + a^2 + b^2$

$722 = 529 + a^2 + b^2$

Rearrange to find $a^2 + b^2$:

$a^2 + b^2 = 722 - 529$

$a^2 + b^2 = 193$ (Equation 2)

Finding the Value of $ab$

We know the identity $(a+b)^2 = a^2 + b^2 + 2ab$.

Substitute the values from Equation 1 ($a+b=19$) and Equation 2 ($a^2+b^2=193$):

$(19)^2 = 193 + 2ab$

$361 = 193 + 2ab$

Solve for $2ab$:

$2ab = 361 - 193$

$2ab = 168$

Solve for $ab$:

$ab = \frac{168}{2}$

$ab = 84$

Calculating the Final Expression

The expression we need to find is $a+b+ab$.

Using the values we found:

• $a+b = 19$
• $ab = 84$

Substitute these values into the expression:

$a+b+ab = 19 + 84 = 103$.