| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Binomial from normal with unknown parameter |
| Difficulty | Standard +0.3 This is a straightforward application of normal distribution with standard transformations to z-scores and one binomial probability calculation. Part (i) is direct standardization, (ii) uses inverse normal with a given percentage, (iii) applies binomial distribution with the 5% from part (ii), and (iv) finds P(T<0). All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(t > 6) = P\!\left(z > \dfrac{6-5.3}{2.1}\right) = P(z > 0.333)\) | M1 | Standardising, no continuity correction, no sq, no sq rt |
| \(= 1 - 0.6304\) | M1 | Correct area \(1 - \Phi\) (\(< 0.5\)), final solution |
| \(= 0.370\) or \(0.369\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(z = 1.645\) | B1 | \(\pm 1.645\) |
| \(1.645 = \dfrac{x - 5.3}{2.1}\) | M1 | Standardising, no continuity correction, allow sq, sq rt. Must be equated to a \(z\)-value |
| \(x = 8.75\) or \(8.755\) or \(8.7545\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(n = 10,\ p = 0.05\) | M1 | Bin term \(^{10}C_x\, p^x(1-p)^{10-x}\) |
| \(P(0,1,2) = (0.95)^{10} + {^{10}C_1}(0.05)(0.95)^9 + {^{10}C_2}(0.05)^2(0.95)^8\) | M1 | Correct unsimplified answer |
| \(= 0.988\) (0.9885 to 4 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{misses bus}) = P(t < 0)\) | *M1 | Seeing \(t\) linked to zero |
| \(= P\!\left(z < \dfrac{0-5.3}{2.1}\right) = P(z < -2.524) = 1 - \Phi(2.524) = 1 - 0.9942\) | DM1 | Standardising with \(t = 0\), no continuity correction, no sq, no sq rt |
| \(= 0.0058\) | A1 |
## Question 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(t > 6) = P\!\left(z > \dfrac{6-5.3}{2.1}\right) = P(z > 0.333)$ | M1 | Standardising, no continuity correction, no sq, no sq rt |
| $= 1 - 0.6304$ | M1 | Correct area $1 - \Phi$ ($< 0.5$), final solution |
| $= 0.370$ or $0.369$ | A1 | |
**Total: 3 marks**
---
## Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $z = 1.645$ | B1 | $\pm 1.645$ |
| $1.645 = \dfrac{x - 5.3}{2.1}$ | M1 | Standardising, no continuity correction, allow sq, sq rt. Must be equated to a $z$-value |
| $x = 8.75$ or $8.755$ or $8.7545$ | A1 | |
**Total: 3 marks**
---
## Question 7(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $n = 10,\ p = 0.05$ | M1 | Bin term $^{10}C_x\, p^x(1-p)^{10-x}$ |
| $P(0,1,2) = (0.95)^{10} + {^{10}C_1}(0.05)(0.95)^9 + {^{10}C_2}(0.05)^2(0.95)^8$ | M1 | Correct unsimplified answer |
| $= 0.988$ (0.9885 to 4 sf) | A1 | |
**Total: 3 marks**
---
## Question 7(iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{misses bus}) = P(t < 0)$ | *M1 | Seeing $t$ linked to zero |
| $= P\!\left(z < \dfrac{0-5.3}{2.1}\right) = P(z < -2.524) = 1 - \Phi(2.524) = 1 - 0.9942$ | DM1 | Standardising with $t = 0$, no continuity correction, no sq, no sq rt |
| $= 0.0058$ | A1 | |
**Total: 3 marks**7 Josie aims to catch a bus which departs at a fixed time every day. Josie arrives at the bus stop $T$ minutes before the bus departs, where $T \sim \mathrm {~N} \left( 5.3,2.1 ^ { 2 } \right)$.\\
(i) Find the probability that Josie has to wait longer than 6 minutes at the bus stop.\\
On $5 \%$ of days Josie has to wait longer than $x$ minutes at the bus stop.\\
(ii) Find the value of $x$.\\
(iii) Find the probability that Josie waits longer than $x$ minutes on fewer than 3 days in 10 days.\\
(iv) Find the probability that Josie misses the bus.\\
\hfill \mbox{\textit{CAIE S1 2017 Q7 [12]}}