1. **Determine the missing probability in the distribution.** Given probabilities: $0.07, 0.20, 0.38, ?, 0.13$ Sum of probabilities must be 1. So, $$0.07 + 0.20 + 0.38 + p(3) + 0.13 = 1$$ Sum known values: $$0.07 + 0.20 + 0.38 + 0.13 = 0.78$$ So, $$p(3) = 1 - 0.78 = 0.22$$ 2. **Check if the distribution is a probability distribution.** Probabilities: $0.05, 0.25, 0.35, 0.25, 0.10$ Check sum: $$0.05 + 0.25 + 0.35 + 0.25 + 0.10 = 1.00$$ All probabilities between 0 and 1 and sum to 1, so this is a valid probability distribution. 3. **Check if the distribution is valid.** Probabilities given: $\frac{3}{4}, \frac{1}{10}, \frac{1}{20}, \frac{1}{25}, \frac{1}{50}, -\frac{1}{100}$ Note the negative probability $-\frac{1}{100}$ which is invalid. Thus this is not a valid probability distribution because probabilities cannot be negative. 4. For the distribution with X from 0 to 8 and given probabilities, calculate mean, variance, and standard deviation. - Mean (Expected value): $$E(X) = \sum x \cdot P(x)$$ $$= 0\times0.02 + 1\times0.02 + 2\times0.06 + 3\times0.06 + 4\times0.08 + 5\times0.22 + 6\times0.30 + 7\times0.16 + 8\times0.08$$ Calculate stepwise: $$= 0 + 0.02 + 0.12 + 0.18 + 0.32 + 1.10 + 1.80 + 1.12 + 0.64 = 5.3$$ - Variance: First find $E(X^2)$: $$E(X^2) = \sum x^2 \cdot P(x)$$ $$= 0^2\times0.02 + 1^2\times0.02 + 2^2\times0.06 + 3^2\times0.06 + 4^2\times0.08 + 5^2\times0.22 + 6^2\times0.30 + 7^2\times0.16 + 8^2\times0.08$$ $$= 0 + 0.02 + 0.24 + 0.54 + 1.28 + 5.50 + 10.8 + 7.84 + 5.12 = 31.34$$ Variance: $$\sigma^2 = E(X^2) - (E(X))^2 = 31.34 - (5.3)^2 = 31.34 - 28.09 = 3.25$$ - Standard deviation: $$\sigma = \sqrt{3.25} \approx 1.803$$ 5. For the distribution with $x = 4,5,6,7,8,9$ and $P(X=x)$ given: \(P(X=4)=\frac{1}{16}, P(X=5)=\frac{1}{16}, P(X=6)=\frac{1}{4}, P(X=7)=\frac{3}{16}, P(X=8)=\frac{1}{8}, P(X=9)=\frac{5}{16}\) - Prove it is a PMF: Sum probabilities: $$\frac{1}{16} + \frac{1}{16} + \frac{1}{4} + \frac{3}{16} + \frac{1}{8} + \frac{5}{16} =$$ Convert all to sixteenths: $$1/16 + 1/16 + 4/16 + 3/16 + 2/16 + 5/16 = 16/16 = 1$$ Hence valid PMF. - Expected value: $$E(X) = \sum x \cdot P(x) = 4\times\frac{1}{16} + 5\times\frac{1}{16} + 6\times\frac{1}{4} + 7\times\frac{3}{16} + 8\times\frac{1}{8} + 9\times\frac{5}{16}$$ Calculate: $$= \frac{4}{16} + \frac{5}{16} + \frac{6}{4} + \frac{21}{16} + \frac{8}{8} + \frac{45}{16}$$ Simplify: $$= 0.25 + 0.3125 + 1.5 + 1.3125 + 1 + 2.8125 = 7.1875$$ - Variance $\sigma^2$: Calculate $E(X^2)$: $$= 4^2\times\frac{1}{16} + 5^2\times\frac{1}{16} + 6^2\times\frac{1}{4} + 7^2\times\frac{3}{16} + 8^2\times\frac{1}{8} + 9^2\times\frac{5}{16}$$ $$= \frac{16}{16} + \frac{25}{16} + \frac{36}{4} + \frac{147}{16} + \frac{64}{8} + \frac{405}{16}$$ Simplify: $$= 1 + 1.5625 + 9 + 9.1875 + 8 + 25.3125 = 54.0625$$ Variance: $$\sigma^2 = E(X^2) - (E(X))^2 = 54.0625 - (7.1875)^2 = 54.0625 - 51.66 = 2.40$$ Standard deviation: $$\sigma = \sqrt{2.40} \approx 1.55$$ - Cumulative distribution function $F(X)$: $$F(4) = \frac{1}{16} = 0.0625$$ $$F(5) = F(4) + \frac{1}{16} = 0.125$$ $$F(6) = F(5) + \frac{1}{4} = 0.125 + 0.25 = 0.375$$ $$F(7) = F(6) + \frac{3}{16} = 0.375 + 0.1875 = 0.5625$$ $$F(8) = F(7) + \frac{1}{8} = 0.5625 + 0.125 = 0.6875$$ $$F(9) = F(8) + \frac{5}{16} = 0.6875 + 0.3125 = 1.0$$ - Probabilities requested: $$P(X > 3) = 1 - F(3) = 1$$ (since $X$ starts at 4) $$P(4 < X < 7) = P(X = 5) + P(X=6) = \frac{1}{16} + \frac{1}{4} = 0.0625 + 0.25 = 0.3125$$ $$P(X=6.5) = 0$$ (not a discrete variable value) - $E(X)$ and $\sigma^2$ already found above. 6. Distribution with X values 0,1,3,4,6 and probabilities $k, 0.3, 0.3, 0.2, 0.1$ - Find $k$: Sum probabilities = 1 $$k + 0.3 + 0.3 + 0.2 + 0.1 = 1$$ $$k + 0.9 = 1 \Rightarrow k = 0.1$$ - Expected value: $$E(X) = 0\times0.1 + 1\times0.3 + 3\times0.3 + 4\times0.2 + 6\times0.1 = 0 + 0.3 + 0.9 + 0.8 + 0.6 = 2.6$$ - CDF: $$F(0) = 0.1$$ $$F(1) = 0.1 + 0.3 = 0.4$$ $$F(3) = 0.4 + 0.3 = 0.7$$ $$F(4) = 0.7 + 0.2 = 0.9$$ $$F(6) = 0.9 + 0.1 = 1$$ - Variance: Calculate $E(X^2)$: $$= 0^2\times0.1 + 1^2\times0.3 + 3^2\times0.3 + 4^2\times0.2 + 6^2\times0.1 = 0 + 0.3 + 2.7 + 3.2 + 3.6 = 9.8$$ Variance: $$\sigma^2 = 9.8 - (2.6)^2 = 9.8 - 6.76 = 3.04$$ Standard deviation: $$\sigma = \sqrt{3.04} \approx 1.744$$ - Probabilities: $$P(X \geq 4) = P(4) + P(6) = 0.2 + 0.1 = 0.3$$ $$P(1 \leq X \leq 4) = P(1) + P(3) + P(4) = 0.3 + 0.3 + 0.2 = 0.8$$ $$P(X=1) = 0.3$$