Math B

Statistical Inference: Practice Problems

From expected values and variances to confidence intervals and tests, practise the core calculations using standard normal table values.

Basic, Standard, Advanced: Grade 11 · Term 3

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Expected value and variance

Formulas

E(X) = \sum x_kp_k,\qquad V(X) = E(X^2) - \{E(X)\}^2,\qquad \sigma(X) = \sqrt{V(X)}

E(aX + b) = aE(X) + b,\qquad V(aX + b) = a^2V(X),\qquad \sigma(aX + b) = |a|\sigma(X)

The binomial distribution

If an event has probability $p$ in one trial, the number $X$ of occurrences in $n$ independent trials follows the binomial distribution $B(n,\ p)$ .

Formulas

P(X = k) = {}_n\mathrm{C}_k\,p^k(1-p)^{n-k},\qquad E(X) = np,\qquad V(X) = np(1 - p)

Normal distributions and standardization

If $X$ follows $N(m,\ \sigma^2)$ , then $Z = \dfrac{X - m}{\sigma}$ follows the standard normal distribution $N(0,\ 1)$ . Probabilities come from the table values $p(z) = P(0 \leqq Z \leqq z)$ and the symmetry of the curve.

Example: for $X \sim N(50,\ 10^2)$ , $P(40 \leqq X \leqq 70) = P(-1 \leqq Z \leqq 2) = 0.3413 + 0.4772 = 0.8185$ .

Approximations

For large $n$ , $B(n,\ p)$ is approximately $N(np,\ np(1-p))$ , and the sample mean $\overline{X}$ is approximately $N\left(m,\ \dfrac{\sigma^2}{n}\right)$ .

Estimation and hypothesis tests

95% confidence intervals

\text{mean: }\ \overline{X} - 1.96\frac{\sigma}{\sqrt{n}} \leqq m \leqq \overline{X} + 1.96\frac{\sigma}{\sqrt{n}}

\text{proportion: }\ R - 1.96\sqrt{\frac{R(1-R)}{n}} \leqq p \leqq R + 1.96\sqrt{\frac{R(1-R)}{n}}

In a hypothesis test, standardize the observed value under the null hypothesis; at the $5\%$ level (two-sided) reject it if $|Z| \geqq 1.96$ .

Q1BasicGr 11 · Term 3Statistical inference

A random variable $X$ has the probability distribution below. Find the expected value $E(X)$ and the variance $V(X)$ .

\begin{array}{c|ccc|c} X & 2 & 3 & 4 & \text{Total} \\ \hline P & \frac{1}{2} & \frac{1}{4} & \frac{1}{4} & 1 \end{array}

(1) $E(X)$
(2) $V(X)$

Answer

(1) $E(X) = \frac{11}{4}$
(2) $V(X) = \frac{11}{16}$

Solution

The expected value is
$E(X) = 2 \cdot \frac{1}{2} + 3 \cdot \frac{1}{4} + 4 \cdot \frac{1}{4} = \frac{11}{4}$
The expected value of $X^2$ is
$E(X^2) = 2^2 \cdot \frac{1}{2} + 3^2 \cdot \frac{1}{4} + 4^2 \cdot \frac{1}{4} = \frac{33}{4}$
Hence
$V(X) = E(X^2) - \{E(X)\}^2 = \frac{33}{4} - \left(\frac{11}{4}\right)^2 = \frac{11}{16}$

Q2StandardGr 11 · Term 3Statistical inference

A random variable $X$ follows the normal distribution $N(20,\ 8^2)$ . Find $P(12 \leqq X \leqq 36)$ . Here $Z$ follows the standard normal distribution $N(0,\ 1)$ , and $P(0 \leqq Z \leqq 1) = 0.3413$ , $P(0 \leqq Z \leqq 2) = 0.4772$ .

Answer

0.8185

Solution

Let $Z = \frac{X - 20}{8}$ ; it follows $N(0,\ 1)$ .
$P(12 \leqq X \leqq 36) = P(-1 \leqq Z \leqq 2)$
Use symmetry and the table values.(where $p(z) = P(0 \leqq Z \leqq z)$ )
$P(-1 \leqq Z \leqq 2) = p(1) + p(2) = 0.8185$

Q3AdvancedGr 11 · Term 3Statistical inference

You want to estimate the support rate $p$ for a policy with $95\%$ confidence; $p$ is expected to be about $0.5$ . How many people must be surveyed so that the width of the confidence interval is at most $0.1$ ?

Answer

385

Solution

The condition on the width is
$2 \times 1.96\sqrt{\frac{0.5 \times 0.5}{n}} \leqq 0.1 \quad\Longrightarrow\quad \sqrt{n} \geqq 19.6$
Squaring,
$n \geqq 384.16 \quad\Longrightarrow\quad n = 385$

Q1BasicGr 11 · Term 3Statistical inference

A bag contains $4$ red balls and $3$ white balls. You draw $3$ balls at the same time. Let $X$ be the number of red balls drawn. Find the expected value and the variance of $X$ .

(1) $E(X)$
(2) $V(X)$

Q2StandardGr 11 · Term 3Statistical inference

Each item is defective with probability $\frac{1}{4}$ . Choosing $192$ items at random, let $X$ be the number of defective items. Using the normal approximation, find $P(54 \leqq X \leqq 60)$ . Here $Z$ follows the standard normal distribution $N(0,\ 1)$ , and $P(0 \leqq Z \leqq 1) = 0.3413$ , $P(0 \leqq Z \leqq 2) = 0.4772$ .

Q3AdvancedGr 11 · Term 3Statistical inference

In a random sample of $400$ households in a region, $80$ were watching a certain program. Find the $95\%$ confidence interval for the viewing rate $p$ in the whole region. Answer in the form $a \leqq p \leqq b$ using decimals.

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