| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Symmetric probability given |
| Difficulty | Moderate -0.8 Part (i) is a direct standardization and table lookup for P(X < 440) with given mean and SD. Part (ii) uses symmetry of the normal distribution to find c from a given symmetric probability (94%), requiring P(Z < c/3.6) = 0.97 and inverse normal lookup. Both parts are routine applications of standard normal distribution techniques with no problem-solving insight required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(x < 440)\) | ||
| \(= P\left(z < \frac{440-445}{3.6}\right) = 1 - \Phi(1.389)\) | M1 | Standardising, no cc, no sq or sq rt |
| \(= 1 - 0.9176\) | M1 | Correct area \((1-\Phi)\) oe (indep) |
| \(\text{Ans} = 0.0824\) | A1 [3] | Rounding to correct answer, accept \(0.0825\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(z = 1.881\) | M1 | \(\pm1.88\) or \(1.881\) or \(1.882\) or \(1.555\) seen \(\pm\) |
| \(\frac{c}{3.6} = 1.881\) | M1 | Equation with \(\pm c/3.6\) or \(2c/3.6\) only \(= z\) or prob (can be implied) |
| \(c = 6.77\) | A1 [3] | Correct answer, accept \(6.78\) |
## Question 3:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(x < 440)$ | | |
| $= P\left(z < \frac{440-445}{3.6}\right) = 1 - \Phi(1.389)$ | M1 | Standardising, no cc, no sq or sq rt |
| $= 1 - 0.9176$ | M1 | Correct area $(1-\Phi)$ oe (indep) |
| $\text{Ans} = 0.0824$ | A1 **[3]** | Rounding to correct answer, accept $0.0825$ |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = 1.881$ | M1 | $\pm1.88$ or $1.881$ or $1.882$ or $1.555$ seen $\pm$ |
| $\frac{c}{3.6} = 1.881$ | M1 | Equation with $\pm c/3.6$ or $2c/3.6$ only $= z$ or prob (can be implied) |
| $c = 6.77$ | A1 **[3]** | Correct answer, accept $6.78$ |
---3 Cans of lemon juice are supposed to contain 440 ml of juice. It is found that the actual volume of juice in a can is normally distributed with mean 445 ml and standard deviation 3.6 ml .\\
(i) Find the probability that a randomly chosen can contains less than 440 ml of juice.
It is found that $94 \%$ of the cans contain between $( 445 - c ) \mathrm { ml }$ and $( 445 + c ) \mathrm { ml }$ of juice.\\
(ii) Find the value of $c$.
\hfill \mbox{\textit{CAIE S1 2013 Q3 [6]}}