1. Interpret the table

   The frequency distribution is:

   x & c & 2c & 3c & 4c & 5c & 6c \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \end{array}$$ So the observations are: - $c$ occurring 2 times - $2c,3c,4c,5c,6c$ each occurring 1 time Total frequency: $$N=2+1+1+1+1+1=7$$

2. Find the mean

   $\bar{x}=\frac{\sum fx}{\sum f}$

   Now, $\sum fx=2(c)+1(2c)+1(3c)+1(4c)+1(5c)+1(6c)$ $=2c+2c+3c+4c+5c+6c=22c$

   Hence, $\bar{x}=\frac{22c}{7}$

3. Find $\sum f x^2$

   $\sum f x^2=2(c^2)+(2c)^2+(3c)^2+(4c)^2+(5c)^2+(6c)^2$ $=2c^2+4c^2+9c^2+16c^2+25c^2+36c^2$ $=92c^2$

4. Use the variance formula

   Variance is $\sigma^2=\frac{\sum f x^2}{N}-\left(\frac{\sum fx}{N}\right)^2$

   Given variance $=160$, so $160=\frac{92c^2}{7}-\left(\frac{22c}{7}\right)^2$

   $160=\frac{92c^2}{7}-\frac{484c^2}{49}$

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

   Therefore, $c^2=49$ $c=7$

   Since $c\in \mathbb{N}$, $c=7$

5. Check options

   • A: $5$ ❌
   • B: $8$ ❌
   • C: $6$ ❌
   • D: $7$ ✅

So the correct answer is Option D.