| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct binomial from normal probability |
| Difficulty | Standard +0.3 Part (i) requires a standard inverse normal calculation using tables or calculator. Part (ii) combines normal probability calculation with binomial distribution (finding P(15<X<16) then using binomial with n=7, k≥2), which is a routine multi-step application of two standard techniques. This is slightly above average difficulty due to the combination of topics, but remains a textbook-style question with no novel insight required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z = -1.406\) | B1 | Rounding to \(\pm 1.41\) seen |
| \(\dfrac{c - 14.2}{3.6} = -1.406\) | M1 | Standardising, allow sq rt no cc |
| \(c = 9.14\) | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P\!\left(\dfrac{15-14.2}{3.6}\right) < z < \left(\dfrac{16-14.2}{3.6}\right)\) | M1 | 2 attempts at standardising, no cc, no sq rt |
| \(= \Phi(0.5) - \Phi(0.222) = 0.6915 - 0.5879\) | M1 | Subt two \(\Phi\)s (indep mark) |
| \(= 0.1036\) | A1 | Needn't be entirely accurate, rounding to 0.10 |
| \(P(\text{at least } 2) = 1 - P(0,1) = 1-(0.8964)^7 - (0.8964)^6(0.1036)\,{}_7C_1\) | M1 | Binomial term with \({}_7C_r p^r(1-p)^{7-r}\) seen \(r \neq 0\), any \(p < 1\) |
| \(= 1 - 0.8413\) | M1 | \(1-P(0)\), \(1-P(1)\), \(1-P(0,1)\) seen their \(p\) |
| \(= 0.159\) | A1 [6] | Correct answer, accept 3sf rounding to 0.16 |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = -1.406$ | B1 | Rounding to $\pm 1.41$ seen |
| $\dfrac{c - 14.2}{3.6} = -1.406$ | M1 | Standardising, allow sq rt no cc |
| $c = 9.14$ | A1 **[3]** | Correct answer |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P\!\left(\dfrac{15-14.2}{3.6}\right) < z < \left(\dfrac{16-14.2}{3.6}\right)$ | M1 | 2 attempts at standardising, no cc, no sq rt |
| $= \Phi(0.5) - \Phi(0.222) = 0.6915 - 0.5879$ | M1 | Subt two $\Phi$s (indep mark) |
| $= 0.1036$ | A1 | Needn't be entirely accurate, rounding to 0.10 |
| $P(\text{at least } 2) = 1 - P(0,1) = 1-(0.8964)^7 - (0.8964)^6(0.1036)\,{}_7C_1$ | M1 | Binomial term with ${}_7C_r p^r(1-p)^{7-r}$ seen $r \neq 0$, any $p < 1$ |
| $= 1 - 0.8413$ | M1 | $1-P(0)$, $1-P(1)$, $1-P(0,1)$ seen their $p$ |
| $= 0.159$ | A1 **[6]** | Correct answer, accept 3sf rounding to 0.16 |
---5 Lengths of a certain type of carrot have a normal distribution with mean 14.2 cm and standard deviation 3.6 cm .\\
(i) $8 \%$ of carrots are shorter than $c \mathrm {~cm}$. Find the value of $c$.\\
(ii) Rebekah picks 7 carrots at random. Find the probability that at least 2 of them have lengths between 15 and 16 cm .
\hfill \mbox{\textit{CAIE S1 2013 Q5 [9]}}