(i) State two properties of a binomial distribution. 1. There is a fixed number of trials, denoted by $n$. 2. Each trial has only two possible outcomes, typically labeled as "success" or "failure". 3. The trials are independent of each other. 4. The probability of success, $p$, is constant for every trial. (ii) A discrete random variable $X$ follows a distribution with parameters $n$ and $p$. Determine the value of $n$ given that the mean and the standard deviation of $X$ are both 0.95. Step 1: Write down the formulas for the mean and standard deviation of a binomial distribution. For a binomial distribution $X \sim B(n, p)$: Mean $E(X) = np$ Variance $Var(X) = np(1-p)$ Standard deviation $\sigma = \sqrt{np(1-p)}$ Step 2: Use the given information to set up equations. Given $E(X) = 0.95$ and $\sigma = 0.95$. $$np = 0.95 \quad (1)$$ $$\sqrt{np(1-p)} = 0.95 \quad (2)$$ Step 3: Solve for $p$ using the standard deviation equation. Square both sides of equation (2): $$np(1-p) = (0.95)^2$$ Substitute $np = 0.95$ from equation (1) into this equation: $$0.95(1-p) = (0.95)^2$$ Divide both sides by 0.95: $$1-p = 0.95$$ $$p = 1 - 0.95$$ $$p = 0.05$$ Step 4: Solve for $n$ using the mean equation and the value of $p$. Substitute $p = 0.05$ into equation (1): $$n(0.05) = 0.95$$ $$n = \frac{0.95}{0.05}$$ $$n = 19$$ The value of $n$ is $\boxed{19}$. (iii) The volume of mineral water Sup bottles is Normally distributed with a mean of 502 ml and standard deviation 2.5 ml. A bottle of mineral water is selected at random. Determine the probability that the bottle will contain Given: $\mu = 502$ ml, $\sigma = 2.5$ ml. To calculate probabilities for a normal distribution, we use the z-score formula: $Z = \frac{X - \mu}{\sigma}$. a) less than 506 ml. Step 1: Calculate the z-score for $X = 506$ ml. $$Z = \frac{506 - 502}{2.5}$$ $$Z = \frac{4}{2.5}$$ $$Z = 1.6$$ Step 2: Find the probability $P(X < 506)$. $$P(X < 506) = P(Z < 1.6)$$ Using a standard normal distribution table, $P(Z < 1.6) = 0.9452$. The probability is $\boxed{0.9452}$. b) between 495 ml and 506 ml. Step 1: Calculate the z-score for $X = 495$ ml. $$Z_1 = \frac{495 - 502}{2.5}$$ $$Z_1 = \frac{-7}{2.5}$$ $$Z_1 = -2.8$$ Step 2: Calculate the z-score for $X = 506$ ml. $$Z_2 = \frac{506 - 502}{2.5}$$ $$Z_2 = \frac{4}{2.5}$$ $$Z_2 = 1.6$$ Step 3: Find the probability $P(495 < X < 506)$. $$P(495 < X < 506) = P(-2.8 < Z < 1.6)$$ $$P(-2.8 < Z < 1.6) = P(Z < 1.6) - P(Z < -2.8)$$ From a standard normal distribution table: $P(Z < 1.6) = 0.9452$ $P(Z < -2.8) = P(Z > 2.8) = 1 - P(Z < 2.8) = 1 - 0.9974 = 0.0026$ $$P(-2.8 < Z < 1.6) = 0.9452 - 0.0026$$ $$P(-2.8 < Z < 1.6) = 0.9426$$ The probability is $\boxed{0.9426}$. c) At least 502 ml Step 1: Calculate the z-score for $X = 502$ ml. $$Z = \frac{502 - 502}{2.5}$$ $$Z = \frac{0}{2.5}$$ $$Z = 0$$ Step 2: Find the probability $P(X \ge 502)$. $$P(X \ge 502) = P(Z \ge 0)$$ Since the normal distribution is symmetric about its mean (where $Z=0$), the probability of being greater than or equal to the mean is 0.5. $$P(Z \ge 0) = 0.5$$ The probability is $\boxed{0.5}$.