| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Moderate -0.3 This is a straightforward normal distribution question requiring standard z-score calculations and expected value. Part (i) and (ii) are routine standardization and inverse normal problems. Part (iii) involves basic probability and expectation but no novel insight—all steps follow directly from the setup. Slightly easier than average due to clear structure and standard techniques. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{small}) = P\!\left(z < \dfrac{95-150}{50}\right)\) | M1 | \(\pm\) standardising using 95, no cc, no sq, no sq rt |
| \(= P(z < -1.1) = 1 - 0.8643 = 0.136\) | M1, A1 [3] | \(1 - \Phi\) (in final answer) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z = 1.282\) | B1 | \(\pm\) rounding to 1.28 |
| \(1.282 = \dfrac{x - 150}{50}\) | M1 | Standardised eqn in their \(z\), allow cc |
| \(x = 214\) g | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{small}) = 0.1357\), \(P(\text{large}) = 0.1357\) symmetry | B1 | Correct answer legit obtained |
| \(P(\text{medium}) = 1 - 0.1357\times2 = 0.7286\) AG | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Expected cost per banana \(= 0.1357\times10 + 0.1357\times25 + 0.7286\times20 = 19.3215\) cents | \*M1 | Attempt at multiplying each 'prob' by a price and summing |
| Total cost of 100 bananas \(= 1930\) (cents) (\$19.30) | DM1, A1 [3] | Mult by 100 |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{small}) = P\!\left(z < \dfrac{95-150}{50}\right)$ | M1 | $\pm$ standardising using 95, no cc, no sq, no sq rt |
| $= P(z < -1.1) = 1 - 0.8643 = 0.136$ | M1, A1 [3] | $1 - \Phi$ (in final answer) |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = 1.282$ | B1 | $\pm$ rounding to 1.28 |
| $1.282 = \dfrac{x - 150}{50}$ | M1 | Standardised eqn in their $z$, allow cc |
| $x = 214$ g | A1 [3] | |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{small}) = 0.1357$, $P(\text{large}) = 0.1357$ symmetry | B1 | Correct answer legit obtained |
| $P(\text{medium}) = 1 - 0.1357\times2 = 0.7286$ AG | [1] | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Expected cost per banana $= 0.1357\times10 + 0.1357\times25 + 0.7286\times20 = 19.3215$ cents | \*M1 | Attempt at multiplying each 'prob' by a price and summing |
| Total cost of 100 bananas $= 1930$ (cents) (\$19.30) | DM1, A1 [3] | Mult by 100 |
---6 The weights of bananas in a fruit shop have a normal distribution with mean 150 grams and standard deviation 50 grams. Three sizes of banana are sold.
Small: under 95 grams\\
Medium: between 95 grams and 205 grams\\
Large: over 205 grams
\begin{enumerate}[label=(\roman*)]
\item Find the proportion of bananas that are small.
\item Find the weight exceeded by $10 \%$ of bananas.
The prices of bananas are 10 cents for a small banana, 20 cents for a medium banana and 25 cents for a large banana.
\item (a) Show that the probability that a randomly chosen banana costs 20 cents is 0.7286 .\\
(b) Calculate the expected total cost of 100 randomly chosen bananas.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2016 Q6 [10]}}