1. **Problem statement:** We have a discrete random variable $X$ with probability mass function (pmf) given by $$P(X=x) = \frac{x+1}{m}, \quad x = 1,2,3,4,5,6.$$ We need to find: i) The value of $m$. ii) The expected value $E[X]$. iii) The standard deviation of $X$. 2. **Step i) Find $m$:** Since $P(X=x)$ is a pmf, the sum of probabilities over all $x$ must be 1: $$\sum_{x=1}^6 P(X=x) = 1 \implies \sum_{x=1}^6 \frac{x+1}{m} = 1.$$ Calculate the sum in the numerator: $$\sum_{x=1}^6 (x+1) = (1+1) + (2+1) + (3+1) + (4+1) + (5+1) + (6+1) = 2 + 3 + 4 + 5 + 6 + 7 = 27.$$ So: $$\frac{27}{m} = 1 \implies m = 27.$$ 3. **Step ii) Calculate $E[X]$:** The expected value is: $$E[X] = \sum_{x=1}^6 x P(X=x) = \sum_{x=1}^6 x \frac{x+1}{27} = \frac{1}{27} \sum_{x=1}^6 x(x+1).$$ Calculate the sum: $$\sum_{x=1}^6 x(x+1) = \sum_{x=1}^6 (x^2 + x) = \sum_{x=1}^6 x^2 + \sum_{x=1}^6 x.$$ Use formulas: $$\sum_{x=1}^6 x = \frac{6 \times 7}{2} = 21,$$ $$\sum_{x=1}^6 x^2 = \frac{6 \times 7 \times 13}{6} = 91.$$ So: $$\sum_{x=1}^6 x(x+1) = 91 + 21 = 112.$$ Therefore: $$E[X] = \frac{112}{27} \approx 4.148.$$ 4. **Step iii) Calculate standard deviation:** First find $E[X^2]$: $$E[X^2] = \sum_{x=1}^6 x^2 P(X=x) = \sum_{x=1}^6 x^2 \frac{x+1}{27} = \frac{1}{27} \sum_{x=1}^6 x^2 (x+1) = \frac{1}{27} \sum_{x=1}^6 (x^3 + x^2).$$ Calculate sums: $$\sum_{x=1}^6 x^3 = \left(\frac{6 \times 7}{2}\right)^2 = 21^2 = 441,$$ $$\sum_{x=1}^6 x^2 = 91.$$ So: $$\sum_{x=1}^6 (x^3 + x^2) = 441 + 91 = 532.$$ Hence: $$E[X^2] = \frac{532}{27} \approx 19.704.$$ Variance is: $$Var(X) = E[X^2] - (E[X])^2 = \frac{532}{27} - \left(\frac{112}{27}\right)^2 = \frac{532}{27} - \frac{12544}{729}.$$ Calculate common denominator 729: $$\frac{532}{27} = \frac{532 \times 27}{27 \times 27} = \frac{14364}{729}.$$ So: $$Var(X) = \frac{14364}{729} - \frac{12544}{729} = \frac{1820}{729} \approx 2.4966.$$ Standard deviation is: $$\sigma = \sqrt{Var(X)} = \sqrt{\frac{1820}{729}} \approx 1.58.$$ **Final answers:** - $m = 27$ - $E[X] = \frac{112}{27} \approx 4.148$ - Standard deviation $\approx 1.58$