| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Standard +0.3 This is a straightforward normal distribution question requiring standard z-score calculations (part i), inverse normal lookup (part ii), and a simple application of independent events with binomial probability (part iii). All techniques are routine for S1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P({<}\,4.5) = P\!\left(z < \dfrac{4.5 - 4.2}{0.6}\right) = P(z < 0.5) = 0.6915\) | M1 | Standardising once, no cc, no sq, no sq rt |
| \(P({<}\,3.5) = P\!\left(z < \dfrac{3.5 - 4.2}{0.6}\right) = P(z < -1.167)\) | ||
| \(= 1 - 0.8784 = 0.1216\) | M1 | \(\Phi_1 - (1 - \Phi_2)\ [P_1 - P_2,\ 1{>}P_1{>}0.5,\ 0.5{>}P_2{>}0]\) oe |
| \(0.6915 - 0.1216 = 0.570\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z = 1.175\) | B1 | \(\pm 1.17\) to 1.18 seen |
| \(1.175 = \dfrac{t - 4.2}{0.6}\) | M1 | Standardising, no cc, allow sq, sq rt with \(z\)-value (not \(\pm\)0.8106, 0.5478, 0.4522, 0.1894, 0.175 etc.) |
| \(t = 4.91\) | A1 [3] | Correct answer from \(z = 1.175\) seen (4sf) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((0.88)^n < 0.003\) | M1 | Inequality or eqn in 0.88, power correctly placed using \(n\) or \((n \pm 1)\), 0.003 or \((1 - 0.003)\) oe |
| \(n > \lg(0.003)/\lg(0.88)\) | M1 | Attempt to solve by logs or trial and error (may be implied by answer) |
| \(n > 45.4\) | ||
| \(n = 46\) | A1 [3] | Correct integer answer |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P({<}\,4.5) = P\!\left(z < \dfrac{4.5 - 4.2}{0.6}\right) = P(z < 0.5) = 0.6915$ | M1 | Standardising once, no cc, no sq, no sq rt |
| $P({<}\,3.5) = P\!\left(z < \dfrac{3.5 - 4.2}{0.6}\right) = P(z < -1.167)$ | | |
| $= 1 - 0.8784 = 0.1216$ | M1 | $\Phi_1 - (1 - \Phi_2)\ [P_1 - P_2,\ 1{>}P_1{>}0.5,\ 0.5{>}P_2{>}0]$ oe |
| $0.6915 - 0.1216 = 0.570$ | A1 [3] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = 1.175$ | B1 | $\pm 1.17$ to 1.18 seen |
| $1.175 = \dfrac{t - 4.2}{0.6}$ | M1 | Standardising, no cc, allow sq, sq rt with $z$-value (not $\pm$0.8106, 0.5478, 0.4522, 0.1894, 0.175 etc.) |
| $t = 4.91$ | A1 [3] | Correct answer from $z = 1.175$ seen (4sf) |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(0.88)^n < 0.003$ | M1 | Inequality or eqn in 0.88, power correctly placed using $n$ or $(n \pm 1)$, 0.003 or $(1 - 0.003)$ oe |
| $n > \lg(0.003)/\lg(0.88)$ | M1 | Attempt to solve by logs or trial and error (may be implied by answer) |
| $n > 45.4$ | | |
| $n = 46$ | A1 [3] | Correct integer answer |
---4 The time taken to cook an egg by people living in a certain town has a normal distribution with mean 4.2 minutes and standard deviation 0.6 minutes.\\
(i) Find the probability that a person chosen at random takes between 3.5 and 4.5 minutes to cook an egg.\\
$12 \%$ of people take more than $t$ minutes to cook an egg.\\
(ii) Find the value of $t$.\\
(iii) A random sample of $n$ people is taken. Find the smallest possible value of $n$ if the probability that none of these people takes more than $t$ minutes to cook an egg is less than 0.003 .
\hfill \mbox{\textit{CAIE S1 2016 Q4 [9]}}