If the variance of the frequency distribution

is 160, then the value of $$c \in \mathbb{N}$$ is

We have a discrete frequency distribution:

• (x = c, 2c, 3c, 4c, 5c, 6c)
• (f = 2, 1, 1, 1, 1, 1)

Total frequency:
N = 2 + 1 + 1 + 1 + 1 + 1 = 7

$$\overline{x}=\frac{\sum_{ }^{ }fx}{N}$$

$$\sum_{ }^{ }fx=2(c)+1(2c)+1(3c)+1(4c)+1(5c)+1(6c)$$

= 2c + 2c + 3c + 4c + 5c + 6c = 22c

$$\overline{x}=\frac{22c}{7}$$

$$\text{Var}=\frac{\sum_{ }^{ }fx^2}{N}-\overline{x}^2$$

Compute:
$$\sum_{ }^{ }fx^2=2c^2+(2c)^2+(3c)^2+(4c)^2+(5c)^2+(6c)^2$$

$$=2c^2+4c^2+9c^2+16c^2+25c^2+36c^2$$

$$=92c^2$$

$$\text{Var}=\frac{92c^2}{7}-\left(\frac{22c}{7}\right)^2$$

$$=\frac{92c^2}{7}-\frac{484c^2}{49}$$

$$=\frac{644c^2-484c^2}{49}$$
$$=\frac{160c^2}{49}$$

$$\frac{160c^2}{49}=160$$

$$c^2=49\Rightarrow c=7$$