| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Geometric distribution probability |
| Difficulty | Standard +0.3 This is a straightforward geometric distribution question with standard calculations (expectation, single probability, cumulative probability) followed by a routine binomial-to-normal approximation. All parts use direct formula application with no novel problem-solving required, making it slightly easier than average for A-level statistics. |
| Spec | 2.04d Normal approximation to binomial5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{\frac{1}{4}} = 4\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{9}{64}\ (= 0.141)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X < 6) = 1 - \left(\frac{3}{4}\right)^5\) | M1 | FT their probability/mean from part (a) |
| \(0.763\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Mean} = 80 \times 0.25 = 20\); \(\text{Var} = 80 \times 0.25 \times 0.75 = 15\) | M1 | |
| \(P(\text{more than } 25) = P\left(z > \frac{25.5 - 20}{\sqrt{15}}\right)\) | M1 | |
| \(P(z > 1.42)\) | M1 | |
| \(1 - 0.9222\) | M1 | |
| \(0.0778\) | A1 |
## Question 5:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{\frac{1}{4}} = 4$ | B1 | |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{9}{64}\ (= 0.141)$ | B1 | |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X < 6) = 1 - \left(\frac{3}{4}\right)^5$ | M1 | FT their probability/mean from part (a) |
| $0.763$ | A1 | |
### Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Mean} = 80 \times 0.25 = 20$; $\text{Var} = 80 \times 0.25 \times 0.75 = 15$ | M1 | |
| $P(\text{more than } 25) = P\left(z > \frac{25.5 - 20}{\sqrt{15}}\right)$ | M1 | |
| $P(z > 1.42)$ | M1 | |
| $1 - 0.9222$ | M1 | |
| $0.0778$ | A1 | |
---5 A pair of fair coins is thrown repeatedly until a pair of tails is obtained. The random variable $X$ denotes the number of throws required to obtain a pair of tails.
\begin{enumerate}[label=(\alph*)]
\item Find the expected value of $X$.
\item Find the probability that exactly 3 throws are required to obtain a pair of tails.
\item Find the probability that fewer than 6 throws are required to obtain a pair of tails.\\
On a different occasion, a pair of fair coins is thrown 80 times.
\item Use an approximation to find the probability that a pair of tails is obtained more than 25 times.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q5 [9]}}