This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions for questions 18, 19, and 20. 18. The random variable X has a normal distribution with mean and standard deviation . The mean and standard deviation are also related by the equation 2 = 5, where ≠ 0. Calculate P(X > 5). Step 1: Express the standard deviation in terms of the mean . Given 2 = 5, we can write: = (5)/(2) = 2.5 Step 2: Standardize the value X = 5 using the Z-score formula. The Z-score formula is Z = (X - )/(). Substitute X = 5 and = 2.5: Z = (5 - )/(2.5) = (4)/(2.5) = (4)/(2.5) = 1.6 Step 3: Calculate the probability P(X > 5), which is equivalent to P(Z > 1.6). Using a standard normal distribution table or calculator, P(Z 1.6) ≈ 0.9452. Therefore, P(Z > 1.6) = 1 - P(Z 1.6): P(Z > 1.6) = 1 - 0.9452 = 0.0548 The correct option is A. The probability P(X > 5) is A. 0.0548. 19. In a Durban suburb, 10% of houses have solar geyser. 16 houses are chosen at random. Find the probability that fewer than 4 houses have solar geyser. Step 1: Identify the parameters for the binomial distribution. This is a binomial distribution problem, where n is the number of trials and p is the probability of success. Number of houses chosen, n = 16. Probability of a house having a solar geyser, p = 0.10. Probability of a house not having a solar geyser, 1-p = 0.90. We need to find the probability that fewer than 4 houses have a solar geyser, which means P(X < 4) or P(X 3). This is P(X=0) + P(X=1) + P(X=2) + P(X=3). Step 2: Use the binomial probability formula P(X=k) = nk p^k (1-p)^n-k for each value of k. For k=0: P(X=0) = 160 (0.10)^0 (0.90)^16 = 1 × 1 × 0.185302 ≈ 0.1853 For k=1: P(X=1) = 161 (0.10)^1 (0.90)^15 = 16 × 0.10 × 0.205891 ≈ 0.3294 For k=2: P(X=2) = 162 (0.10)^2 (0.90)^14 = 120 × 0.01 × 0.228768 ≈ 0.2745 For k=3: P(X=3) = 163 (0.10)^3 (0.90)^13 = 560 × 0.001 × 0.254187 ≈ 0.1423 Step 3: Sum the probabilities. P(X < 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) P(X < 4) = 0.1853 + 0.3294 + 0.2745 + 0.1423 = 0.9315 Rounding to four decimal places, $P(X < 4) ≈ 0.931