If a,b,7,10,11,15, mean=10, variance=`20/3` and variance is then find value of a and b

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) given the numbers \( a, b, 7, 10, 11, 15 \) with a mean of \( 10 \) and a variance of \( \frac{20}{3} \).

### Step 1: Calculate the Mean
The formula for the mean (\( \bar{x} \)) is given by:

\[ \bar{x} = \frac{\text{sum of all observations}}{n} \]

Here, we have 6 observations: \( a, b, 7, 10, 11, 15 \). The mean is given as \( 10 \).

Setting up the equation:

\[ 10 = \frac{a + b + 7 + 10 + 11 + 15}{6} \]

Calculating the sum of the known values:

\[ 10 = \frac{a + b + 43}{6} \]

Cross-multiplying gives:

\[ 60 = a + b + 43 \]

Rearranging this, we find:

\[ a + b = 60 - 43 = 17 \quad \text{(Equation 1)} \]

### Step 2: Calculate the Variance
The formula for variance (\( \sigma^2 \)) is given by:

\[ \sigma^2 = \frac{\sum x^2}{n} - \left(\bar{x}\right)^2 \]

We know the variance is \( \frac{20}{3} \) and the mean is \( 10 \). Plugging in the values:

\[ \frac{20}{3} = \frac{a^2 + b^2 + 7^2 + 10^2 + 11^2 + 15^2}{6} - 10^2 \]

Calculating the squares of the known values:

\[ 7^2 = 49, \quad 10^2 = 100, \quad 11^2 = 121, \quad 15^2 = 225 \]

Thus, we have:

\[ \frac{20}{3} = \frac{a^2 + b^2 + 49 + 100 + 121 + 225}{6} - 100 \]

Calculating the sum of squares:

\[ 49 + 100 + 121 + 225 = 495 \]

Substituting back into the equation:

\[ \frac{20}{3} = \frac{a^2 + b^2 + 495}{6} - 100 \]

Multiplying through by \( 6 \):

\[ 40 = a^2 + b^2 + 495 - 600 \]

This simplifies to:

\[ a^2 + b^2 - 105 = 40 \]

Rearranging gives us:

\[ a^2 + b^2 = 145 \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations
Now we have two equations:

1. \( a + b = 17 \)
2. \( a^2 + b^2 = 145 \)

From Equation 1, we can express \( a \) in terms of \( b \):

\[ a = 17 - b \]

Substituting this into Equation 2:

\[ (17 - b)^2 + b^2 = 145 \]

Expanding the left side:

\[ 289 - 34b + b^2 + b^2 = 145 \]

This simplifies to:

\[ 2b^2 - 34b + 289 - 145 = 0 \]

Which simplifies further to:

\[ 2b^2 - 34b + 144 = 0 \]

Dividing the entire equation by 2 gives:

\[ b^2 - 17b + 72 = 0 \]

### Step 4: Factor the Quadratic Equation
Factoring the quadratic:

\[ (b - 8)(b - 9) = 0 \]

Thus, we find:

\[ b = 8 \quad \text{or} \quad b = 9 \]

### Step 5: Find Corresponding Values of \( a \)
Using \( b = 8 \):

\[ a = 17 - 8 = 9 \]

Using \( b = 9 \):

\[ a = 17 - 9 = 8 \]

### Final Solution
Thus, the values of \( a \) and \( b \) are:

\[ (a, b) = (9, 8) \quad \text{or} \quad (8, 9) \]