From the image provided, it appears to be a statistics problem where the mean is given as 50, and you are tasked with determining the missing frequencies $f_1$ and $f_2$. The problem also involves a frequency distribution table with the following class intervals and frequencies:

Class Interval (C.I.)	Frequency (f)
10–30	$f_1$
30–50	90
50–70	30
70–90	$f_2$
90–110	40
Total	200

The mean formula for a grouped frequency distribution is:

$\text{Mean} = \frac{\sum f \cdot x}{\sum f}$

Here:

• $f$ represents frequency.
• $x$ represents the class mark of each interval, calculated as: $x = \frac{\text{Lower Bound} + \text{Upper Bound}}{2}$
• Total frequency $\sum f = 200$, and the mean is given as 50.

Step 1: Calculate Class Marks ($x$)

C.I.	Class Mark ($x$)
10–30	$x = \frac{10+30}{2} = 20$
30–50	$x = \frac{30+50}{2} = 40$
50–70	$x = \frac{50+70}{2} = 60$
70–90	$x = \frac{70+90}{2} = 80$
90–110	$x = \frac{90+110}{2} = 100$

Step 2: Expand the Formula

Substitute values into the mean formula:

$\text{Mean} = 50 = \frac{\sum f \cdot x}{200}$

Substituting into the formula:

$50 = \frac{(f_1 \cdot 20) + (90 \cdot 40) + (30 \cdot 60) + (f_2 \cdot 80) + (40 \cdot 100)}{200}$

Simplify the terms:

$50 = \frac{20f_1 + 3600 + 1800 + 80f_2 + 4000}{200}$

Combine like terms:

$50 = \frac{20f_1 + 80f_2 + 9400}{200}$

Multiply through by 200:

$10000 = 20f_1 + 80f_2 + 9400$

Simplify further:

$20f_1 + 80f_2 = 600$

Step 3: Use the Total Frequency Condition

The sum of frequencies is given as 200:

$f_1 + 90 + 30 + f_2 + 40 = 200$

Simplify:

$f_1 + f_2 + 160 = 200$

$f_1 + f_2 = 40$

Step 4: Solve the System of Equations

We now have two equations:

1. $20f_1 + 80f_2 = 600$
2. $f_1 + f_2 = 40$

From the second equation, solve for $f_1$: $f_1 = 40 - f_2$

Substitute $f_1 = 40 - f_2$ into the first equation:

$20(40 - f_2) + 80f_2 = 600$

Expand:

$800 - 20f_2 + 80f_2 = 600$

Simplify:

$800 + 60f_2 = 600$

$60f_2 = -200$

From the equations: 1. $$20f_1 + 80f_2 = 600$$ 2. $$f_1 + f_2 = 40$$ ### Substitute $$f_1 = 40 - f_2$$ into Equation 1: \[ 20(40 - f_2) + 80f_2 = 600$$ Expand: $$800 - 20f_2 + 80f_2 = 600$$ Simplify: $$800 + 60f_2 = 600$$ $$60f_2 = 600 - 800$$ $$60f_2 = -200$$ \[ f_2 = \frac{-200}{60} = -3.33