| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2003 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Exact binomial then normal approximation (same context, different n) |
| Difficulty | Moderate -0.3 This is a straightforward application of binomial distribution followed by normal approximation. Part (i) requires direct calculation of P(X=24) for B(30,0.7), and part (ii) involves standard normal approximation with continuity correction. Both are routine textbook exercises requiring no problem-solving insight, though the calculations involve multiple steps and proper application of the continuity correction. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((0.7)^{24} \times (0.3)^6 \times {}_{30}C_{24}\) | M1 | For relevant binomial calculation |
| \(= 0.0829\) | A1 (2) | For correct answer |
| OR normal approx: \(P(24) = \Phi\!\left(\frac{24.5-21}{\sqrt{6.3}}\right) - \Phi\!\left(\frac{23.5-21}{\sqrt{6.3}}\right)\) | M1 | For subtracting the 2 phi values as written |
| \(= 0.9183 - 0.8404 = 0.0779\) | A1 (2) | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\mu = 30 \times 0.7 = 21\), \(\sigma^2 = 30 \times 0.7 \times 0.3 = 6.3\) | B1 | For 21 and 6.3 seen |
| \(P(<20) = \Phi\!\left(\dfrac{19.5 - 21}{\sqrt{6.3}}\right)\) | M1 | For standardising process, must have \(\sqrt{\,}\), can be \(+\) or \(-\) |
| \(\Phi(-0.5976)\) | M1 | For continuity correction 19.5 or 20.5 |
| M1 | For using \(1 -\) some area found from tables | |
| \(= 1 - 0.7251 = 0.275\) | A1 (5) | For correct answer |
## Question 4:
**(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(0.7)^{24} \times (0.3)^6 \times {}_{30}C_{24}$ | M1 | For relevant binomial calculation |
| $= 0.0829$ | A1 (2) | For correct answer |
| **OR** normal approx: $P(24) = \Phi\!\left(\frac{24.5-21}{\sqrt{6.3}}\right) - \Phi\!\left(\frac{23.5-21}{\sqrt{6.3}}\right)$ | M1 | For subtracting the 2 phi values as written |
| $= 0.9183 - 0.8404 = 0.0779$ | A1 (2) | For correct answer |
**(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mu = 30 \times 0.7 = 21$, $\sigma^2 = 30 \times 0.7 \times 0.3 = 6.3$ | B1 | For 21 and 6.3 seen |
| $P(<20) = \Phi\!\left(\dfrac{19.5 - 21}{\sqrt{6.3}}\right)$ | M1 | For standardising process, must have $\sqrt{\,}$, can be $+$ or $-$ |
| $\Phi(-0.5976)$ | M1 | For continuity correction 19.5 or 20.5 |
| | M1 | For using $1 -$ some area found from tables |
| $= 1 - 0.7251 = 0.275$ | A1 (5) | For correct answer |
---4 Kamal has 30 hens. The probability that any hen lays an egg on any day is 0.7 . Hens do not lay more than one egg per day, and the days on which a hen lays an egg are independent.\\
(i) Calculate the probability that, on any particular day, Kamal's hens lay exactly 24 eggs.\\
(ii) Use a suitable approximation to calculate the probability that Kamal's hens lay fewer than 20 eggs on any particular day.
\hfill \mbox{\textit{CAIE S1 2003 Q4 [7]}}