The table is given to find the random variance.
To find random variance, we have to find the mean,
The mean is 2.5 and the variance is 0.16875.
The variance of a discrete random variable X measures the spread, or variability, of the distribution. Also, the random variance is used to find the standard deviation which is the suare root of the random variance.
so, the answer for standard deviation is 0.412 that is 0.3 is the required answer.
[tex]\begin{gathered} \text{The mean is given by, }\mu_x=\sum ^{\infty}_{n\mathop=0}x_ip_i \\ so,\mu_x=(0\cdot0.027)+(1\cdot0.155)+(2\cdot0.318)+(3\cdot0.318_{})+(4\cdot0.155)+(5\cdot0.027) \\ =0+0.155+0.636+0.954+0.62+0.135=2.5 \\ \text{Hence, }\mu_x=2.5. \\ \text{now, calculate random variance as follows:} \\ \text{Random variance is given by, }\sigma^2_x=\sum ^{\infty}_{n\mathop{=}0}(x_i-\mu_x)p_i\text{.} \\ \text{Calculation part:} \\ \text{ }\sigma^2_x=(0-2.5)(0)+(1-2.5)(0.155)+(2-2.5)(0.318)+(3-2.5)(0.318)+(4-2.5)(0.155)+(5-2.5)(0.027) \\ =0+(-1.5)(0.155)+(-0.5)(0.318)+(0.5)(0.318)+(1.5)(0.155)+(2.5)(0.027) \\ =0-0.2325-0.159+0.159+0.02325+0.0675 \\ =0.0675 \\ \text{Thu, the required random variance is }\sigma^2_x=0.0675. \\ The\text{ meaning of random variance is usedd to find the discrete }random\text{ variable X meansures the spread or variability of the distribution and it is defined as }\sigma^2_x=\sum ^{\infty}_{n\mathop{=}0}(x_i-\mu_x)p_i\text{.} \\ Also,\text{ the standard deviation of }\sigma\text{ is the square root of the variance.} \end{gathered}[/tex]