Covers Lessons 11–15: random variables, probability functions, cumulative distribution functions, the normal distribution, normal applications, the binomial distribution, and module synthesis.
Assessment
Select the best answer for each question.
Which of the following is a continuous random variable?
For a valid probability function $p(x)$, which property must hold?
The heights of adult males are $N(175, 64)$ cm. What is the standard deviation?
Approximately what percentage of values in a normal distribution lie within $\mu \pm 2\sigma$?
A student scores 84 on a test where $\mu = 72$ and $\sigma = 12$. What is their z-score?
Which condition is NOT required for a binomial distribution?
For $X \sim B(20, 0.3)$, what is $E(X)$?
When is normal approximation to the binomial appropriate?
Short Answer
A discrete random variable $X$ has the following probability function:
| $x$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $p(x)$ | 0.10 | 0.25 | $k$ | 0.30 | 0.15 |
(a) Find the value of $k$. (b) Find $P(X \leq 2)$. (c) Calculate $E(X)$. 3 MARKS
The weights of apples from an orchard are normally distributed with $\mu = 150$ g and $\sigma = 15$ g.
(a) Calculate the z-score for an apple weighing 180 g. (b) Use the empirical rule to estimate the percentage of apples weighing between 120 g and 180 g. (c) Find the weight below which $97.5\%$ of apples fall. 3 MARKS
A quality control process tests 12 items from a large batch. Historically, $8\%$ of items are defective. Defects occur independently.
(a) Name the distribution that models the number of defective items and state its parameters. (b) Find the probability that exactly one item is defective. (c) A manager wants to use normal approximation to estimate $P(X \leq 2)$. Evaluate whether this is appropriate and, if so, state the approximating normal distribution. 3 MARKS
Q1: B — Height can take any value in a continuum. The others are countable (discrete).
Q2: B — A valid probability function requires $0 \leq p(x) \leq 1$ for all $x$ and total probability 1.
Q3: B — $N(\mu, \sigma^2)$ means variance = 64, so $\sigma = \sqrt{64} = 8$ cm.
Q4: B — The empirical rule states approximately 95% within $\mu \pm 2\sigma$.
Q5: C — $z = \frac{84 - 72}{12} = \frac{12}{12} = 1.0$.
Q6: C — The binomial requires exactly two outcomes (success/failure), not three.
Q7: A — $E(X) = np = 20 \times 0.3 = 6$.
Q8: B — Both $np \geq 5$ and $n(1-p) \geq 5$ are required for valid normal approximation.
Q9 (3 marks): (a) $0.10 + 0.25 + k + 0.30 + 0.15 = 1$, so $k = 0.20$ [1]. (b) $P(X \leq 2) = 0.10 + 0.25 + 0.20 = 0.55$ [1]. (c) $E(X) = 0(0.10) + 1(0.25) + 2(0.20) + 3(0.30) + 4(0.15) = 0 + 0.25 + 0.40 + 0.90 + 0.60 = 2.15$ [1].
Q10 (3 marks): (a) $z = \frac{180 - 150}{15} = 2$ [0.5]. (b) $120 = \mu - 2\sigma$ and $180 = \mu + 2\sigma$. By the empirical rule, approximately $95\%$ of apples weigh between 120 g and 180 g [1 + 0.5]. (c) $97.5\%$ corresponds to $\mu + 2\sigma = 150 + 30 = 180$ g [1].
Q11 (3 marks): (a) $X \sim B(12, 0.08)$ — binomial with $n = 12$ and $p = 0.08$ [0.5 + 0.5]. (b) $P(X = 1) = \binom{12}{1}(0.08)^1(0.92)^{11} = 12 \times 0.08 \times 0.3996 \approx 0.384$ [1]. (c) $np = 12 \times 0.08 = 0.96 < 5$ [0.5]. Since $np < 5$, normal approximation is NOT appropriate — the distribution is too skewed [0.5].