Frequency Distribution Data

The problem provides a frequency distribution with class intervals and frequencies. We need to find the values of the unknown frequency $\lambda$ and the mean $\mu$ using the given variance.

Calculate Sums

First, calculate the total frequency and the sums needed for the mean and variance formulas.

• Total Frequency: $\sum f_i = 3 + \lambda + 4 + 7 = 14 + \lambda$
• Sum of $f_i x_i$: $\sum f_i x_i = 18 + 10\lambda + 56 + 126 = 200 + 10\lambda$
• Sum of $f_i x_i^2$: $\sum f_i x_i^2 = 108 + 100\lambda + 784 + 2268 = 3160 + 100\lambda$

Mean Calculation

The formula for the mean ($\mu$) of a grouped frequency distribution is:

$ \mu = \frac{\sum f_i x_i}{\sum f_i} $

Substituting the sums:

$ \mu = \frac{200 + 10\lambda}{14 + \lambda} $ (Equation 1)

Variance Calculation

The formula for the variance ($\sigma^2$) is:

$ \sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - \mu^2 $

We are given that the variance is 19:

$ 19 = \frac{3160 + 100\lambda}{14 + \lambda} - \mu^2 $ (Equation 2)

Solve for $\lambda$ and $\mu$

Substitute Equation 1 into Equation 2:

$ 19 = \frac{3160 + 100\lambda}{14 + \lambda} - \left( \frac{200 + 10\lambda}{14 + \lambda} \right)^2 $

Multiply by $(14 + \lambda)^2$ to clear denominators:

$ 19(14 + \lambda)^2 = (3160 + 100\lambda)(14 + \lambda) - (200 + 10\lambda)^2 $

Expand and simplify the equation:

$ 19(196 + 28\lambda + \lambda^2) = (44240 + 3160\lambda + 1400\lambda + 100\lambda^2) - (40000 + 4000\lambda + 100\lambda^2) $

$ 3724 + 532\lambda + 19\lambda^2 = 4240 + 560\lambda $

Rearrange into a quadratic equation for $\lambda$:

$ 19\lambda^2 - 28\lambda - 516 = 0 $

Solve the quadratic equation using the formula $\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$ \lambda = \frac{28 \pm \sqrt{(-28)^2 - 4(19)(-516)}}{2(19)} = \frac{28 \pm \sqrt{784 + 39216}}{38} = \frac{28 \pm \sqrt{40000}}{38} = \frac{28 \pm 200}{38} $

Since frequency must be positive, we take the positive root:

$ \lambda = \frac{28 + 200}{38} = \frac{228}{38} = 6 $

Now, find the mean $\mu$ using Equation 1 with $\lambda = 6$:

$ \mu = \frac{200 + 10(6)}{14 + 6} = \frac{200 + 60}{20} = \frac{260}{20} = 13 $

We have found $\lambda = 6$ and $\mu = 13$. Verifying the variance:

$ \sigma^2 = \frac{3160 + 100(6)}{14 + 6} - 13^2 = \frac{3760}{20} - 169 = 188 - 169 = 19 $

The calculated variance matches the given value.

Final Calculation

The question asks for the value of $\lambda + \mu$:

$ \lambda + \mu = 6 + 13 = 19 $