# Question 7:

## Part (i)(A)

$X\sim B(16, 0.1)$

$P(X=3) = 0.25^4\times0.75^{16}\binom{20}{4}\times0.25^4\times0.75^{16} = 0.1423$ | M1 | For $0.1^3\times0.9^{13}$

| M1 | For $\binom{16}{3}\times p^3\times q^{13}$

| A1 [3] | CAO

Or from tables: $P(X\leq3) - P(X\leq2) = 0.9316 - 0.7892 = 0.1424$ | M2, A1 | For $0.9316 - 0.7892$

---

## Part (i)(B)

$P(X\geq3) = 1 - P(X\leq2) = 1 - 0.7892 = 0.2108$ | M1 | For 0.7892

| A1 [2] | CAO; if calculating $P(X=0)+P(X=1)+P(X=2)$ allow M1 for 0.79 or better

---

## Part (i)(C)

Expected number $= 16\times0.1 = 1.6$ | B1 [1] | Do not allow final answer of 1 or 2 even if correct 1.6 given earlier

---

## Part (ii)

Let $p$ = probability of a randomly chosen person using 1234 as their PIN (in the population) | B1 | For definition of $p$ in context; do NOT allow number in place of probability

$H_0: p = 0.1$ | B1 | For $H_0$

$H_1: p < 0.1$ | B1 |

The alternative hypothesis has this form as the advertising campaign aims to reduce the proportion of the population who use 1234 as their PIN | B1 [4] | Dep on $< 0.1$ used in $H_1$; do not allow just 'proportion will be lower' or similar

---

## Part (iii)(A)

For $n=20$, $P(X\leq0) = 0.1216$ | M1* | For sight of 0.1216; condone $P(X=0)$ in place of $P(X\leq0)$

$0.1216 > 0.10$ | *M1dep | For $> 0.10$ or $> 10\%$; do NOT FT wrong $H_1$

So no point in carrying out the test as $H_0$ could not be rejected (even if nobody in the sample uses 1234 as their PIN) oe | A1 [3] | or state 'There is no critical region' oe; for A1 need $P(X\leq0)$ or $P(X=0)$ somewhere

---

## Part (iii)(B)

Lowest value of $k$ is 13 | B1 [1] | Or 13%

---

## Part (iv)

$P(X\leq2) = 0.0530$ | B1 | For use of $P(X\leq2)$ only

$0.0530 > 0.05$ | M1 | For comparison of 0.0530 with 5%

So not significant. Do not reject $H_0$ | A1* |

Conclude that there is not enough evidence to support the suggestion that the advertising campaign has been successful | *E1 dep [4] | Must include 'insufficient evidence to suggest that' or similar; allow 'accept $H_0$' or 'reject $H_1$'

## Additional Notes: Q7 Part (ii) — Hypotheses for Binomial Test

---

**Definitions and Allowances for p (B1):**

| Guidance |
|----------|
| Minimum needed for B1: $p$ = probability of using 1234 |
| Allow $p = P(\text{using } 1234)$ |
| Definition must include word *probability* (or *chance*, *proportion*, *percentage*, *likelihood*) — NOT *possibility*, *number*, or *amount* |
| Preferably given as separate comment, but can appear at end of $H_0$ if clearly stated as '$p$ = the probability of using 1234' |
| Do NOT allow '$p$ = the probability of using 1234 is different' |
| Allow $p = 10\%$; allow only $p$, $\theta$, $\pi$, or $\rho$; allow any single symbol **if defined** (including $x$) |
| Allow $H_0 = p = 0.1$; Allow $H_0 : p = \frac{1}{10}$ |
| Allow NH and AH in place of $H_0$ and $H_1$ |
| Do not allow $H_0 : P(X = x) = 0.1$ |
| Do not allow $H_0: {=}0.1,\ {=}10\%,\ P(0.1),\ p(x){=}0.1,\ x{=}0.1$ (unless $x$ correctly defined as a probability) |
| Do not allow $H_0$ and $H_1$ reversed |
| For hypotheses given in words: allow Maximum **B0B1B1** |
| Hypotheses in words must include *probability* (or *chance*, *proportion*, *percentage*) and the figure $0.1$ |
| e.g. $H_0: P(\text{using } 1234) = 0.1,\ H_1: P(\text{using } 1234) < 0.1$ gets **B0B1B1** |