The mean of the following distribution is 31. 4. Determine the missing frequency 𝑥.

Class	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60
Frequency	5	x	10	12	7	8

We are given the mean of the given data, which is 31. 4.

The mean can be found by taking the ratio of the sum of the product of the class mark and the frequency in each case and the total number of frequencies

Here, the total product of the class mark and the frequency, $\sum _{ }{f}_i{x}_i= 1450 + 15x$

And the total number of families, Σ𝑓𝑖 = 42 + 𝑥

So, mean =$\frac{{\displaystyle \sum _{ }}{f}_i{x}_i}{{\displaystyle \sum _{ }}{f}_i}$

31.4 = $\frac{{\displaystyle 1450+15x}}{{\displaystyle 42+x}}$

⇒ 31. 4(42 + 𝑥) = 1450 + 15𝑥

⇒ 1318. 8 + 31. 4𝑥 = 1450 + 15𝑥

⇒ 31. 4𝑥 − 15𝑥 = 1450 − 1318. 8

⇒ 16. 4𝑥 = 131. 2
x = 131.2 /16.4
⇒ 𝑥 = 8
Hence, the value of 𝑥 is 8