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The mean of the following frequency distribution is 48.2. Two frequencies are missing. Find the missing frequencies. [Table of data]

April 27, 2026Mathematics

ScanSolve AI Answer

Answer

f_A = 7, f_B = 3

Step-by-step solution

4 steps

To find the two missing frequencies, we will use the formula for the arithmetic mean of grouped data and the total number of observations.

The formula for the arithmetic mean ( $\bar{x}$ ) is:

\bar{x} = \frac{\sum f_i x_i}{\sum f_i}

where $f_i$ is the frequency of each class and $x_i$ is the midpoint of each class.

Calculate the midpoint ( $x_i$ ) for each class interval.
- For 20-30: $x_1 = \frac{20+30}{2} = 25$
- For 30-40: $x_2 = \frac{30+40}{2} = 35$
- For 40-50: $x_3 = \frac{40+50}{2} = 45$
- For 50-60: $x_4 = \frac{50+60}{2} = 55$
- For 60-70: $x_5 = \frac{60+70}{2} = 65$

Set up a table with the given frequencies and calculated midpoints.

Let the missing frequencies be $f_A$ for the 30-40 class and $f_B$ for the 60-70 class.

Class interval	Frequency ( $f_i$ )	Midpoint ( $x_i$ )	$f_i x_i$
20-30	3	25	$3 \times 25 = 75$
30-40	$f_A$	35	$35 f_A$
40-50	10	45	$10 \times 45 = 450$
50-60	7	55	$7 \times 55 = 385$
60-70	$f_B$	65	$65 f_B$
Total	$\sum f_i$		$\sum f_i x_i$

Formulate equations based on the given total number of observations and the arithmetic mean.
Equation 1: Sum of frequencies
The total number of observations is given as 30.
$\sum f_i = 3 + f_A + 10 + 7 + f_B = 30$ $20 + f_A + f_B = 30$ $f_A + f_B = 30 - 20$ $f_A + f_B = 10 \quad (1)$
Equation 2: Arithmetic mean
The arithmetic mean is given as 45.
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$ $45 = \frac{75 + 35 f_A + 450 + 385 + 65 f_B}{30}$ $45 \times 30 = 75 + 35 f_A + 450 + 385 + 65 f_B$ $1350 = 910 + 35 f_A + 65 f_B$ $1350 - 910 = 35 f_A + 65 f_B$ $440 = 35 f_A + 65 f_B$
Divide the equation by 5 to simplify:
$\frac{440}{5} = \frac{35 f_A}{5} + \frac{65 f_B}{5}$ $88 = 7 f_A + 13 f_B \quad (2)$
Solve the system of linear equations.
From equation (1), express $f_A$ in terms of $f_B$ :
$f_A = 10 - f_B$
Substitute this into equation (2):
$88 = 7(10 - f_B) + 13 f_B$ $88 = 70 - 7 f_B + 13 f_B$ $88 - 70 = 6 f_B$ $18 = 6 f_B$ $f_B = \frac{18}{6}$ $f_B = 3$
Now substitute the value of $f_B$ back into equation (1) to find $f_A$ :
$f_A + 3 = 10$ $f_A = 10 - 3$ $f_A = 7$
The two missing frequencies are $f_A = 7, f_B = 3$ .

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