1. **Problem statement:** (i) Given the probability density function (pdf) of a continuous random variable $X$: $$f(x) = \begin{cases} \frac{x^3}{20} & 1 \leq x \leq 3 \\ 0 & \text{otherwise} \end{cases}$$ Find (a) $E(X^2)$ using algebraic integration, and (b) the variance $\mathrm{Var}(X)$ given $E(X) = 2.42$. (ii) For a random sample of size 10 from a continuous random variable $Y$: (a) Find the probability that at least 7 values are smaller than the upper quartile of $Y$. (b) Find the probability that less than or equal to 5 values are larger than the upper quartile of $Y$. --- 2. **Step (i)(a): Find $E(X^2)$** The expectation of $X^2$ is given by the integral: $$E(X^2) = \int_{-\infty}^{\infty} x^2 f(x) \, dx = \int_1^3 x^2 \cdot \frac{x^3}{20} \, dx = \int_1^3 \frac{x^5}{20} \, dx$$ Calculate the integral: $$E(X^2) = \frac{1}{20} \int_1^3 x^5 \, dx = \frac{1}{20} \left[ \frac{x^6}{6} \right]_1^3 = \frac{1}{20} \left( \frac{3^6}{6} - \frac{1^6}{6} \right) = \frac{1}{20} \cdot \frac{729 - 1}{6} = \frac{728}{120} = 6.0667$$ --- 3. **Step (i)(b): Find $\mathrm{Var}(X)$** Recall variance formula: $$\mathrm{Var}(X) = E(X^2) - [E(X)]^2$$ Given $E(X) = 2.42$, substitute values: $$\mathrm{Var}(X) = 6.0667 - (2.42)^2 = 6.0667 - 5.8564 = 0.2103$$ Rounded to 3 significant figures: $$\mathrm{Var}(X) = 0.210$$ --- 4. **Step (ii)(a): Probability at least 7 values are smaller than the upper quartile of $Y$** The upper quartile corresponds to the 75th percentile, so the probability a single value is smaller than the upper quartile is $p = 0.75$. Let $X$ be the number of values smaller than the upper quartile in a sample of size 10. Then $X \sim \mathrm{Binomial}(n=10, p=0.75)$. We want: $$P(X \geq 7) = \sum_{k=7}^{10} \binom{10}{k} (0.75)^k (0.25)^{10-k}$$ Calculate each term: - $\binom{10}{7} (0.75)^7 (0.25)^3 = 120 \times 0.1335 \times 0.0156 = 0.2503$ - $\binom{10}{8} (0.75)^8 (0.25)^2 = 45 \times 0.1001 \times 0.0625 = 0.2815$ - $\binom{10}{9} (0.75)^9 (0.25)^1 = 10 \times 0.0751 \times 0.25 = 0.1877$ - $\binom{10}{10} (0.75)^{10} (0.25)^0 = 1 \times 0.0563 \times 1 = 0.0563$ Sum: $$P(X \geq 7) = 0.2503 + 0.2815 + 0.1877 + 0.0563 = 0.7758$$ --- 5. **Step (ii)(b): Probability less than or equal to 5 values are larger than the upper quartile** The probability a single value is larger than the upper quartile is $q = 1 - 0.75 = 0.25$. Let $Y$ be the number of values larger than the upper quartile in the sample. Then $Y \sim \mathrm{Binomial}(n=10, p=0.25)$. We want: $$P(Y \leq 5) = \sum_{k=0}^5 \binom{10}{k} (0.25)^k (0.75)^{10-k}$$ Calculate this cumulative probability (using binomial tables or software) gives approximately: $$P(Y \leq 5) = 0.9939$$ --- **Final answers:** - (i)(a) $E(X^2) = 6.067$ - (i)(b) $\mathrm{Var}(X) = 0.210$ - (ii)(a) $P(\text{at least 7 smaller than upper quartile}) = 0.776$ - (ii)(b) $P(\text{at most 5 larger than upper quartile}) = 0.994$