| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | March |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find median or percentiles |
| Difficulty | Standard +0.3 This is a standard continuous probability distribution question requiring integration to find probabilities and percentiles. Part (a) is routine integration, part (b) involves setting up and simplifying an equation (the algebra is straightforward), and part (c) uses symmetry of the pdf. All techniques are standard for S2 level with no novel insights required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{3}{4000}\int_5^{10}(100 - x^2)\,dx = \frac{3}{4000}\left[100x - \frac{x^3}{3}\right]_5^{10}\) | M1 | Attempt integration of \(f(x)\), ignore limits. Condone omission of \(\frac{3}{4000}\) |
| \(= \frac{3}{4000}\left(1000 - \frac{1000}{3} - 500 + \frac{125}{3}\right)\) | M1 | Correct limits 5 and 10. OE SOI |
| \(= 0.156\) (3 sf) or \(\frac{5}{32}\) | A1 | For fully correct working seen including substitution of limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{3}{4000}\int_p^{10}(100 - x^2)\,dx = \frac{1}{4}\) | M1 | Attempt integration of \(f(x)\) with any limits and \(= \frac{1}{4}\) or \(= \frac{3}{4}\) seen. Condone omission of \(\frac{3}{4000}\) |
| \(\frac{3}{4000}\left[100x - \frac{x^3}{3}\right]_p^{10} = \frac{1}{4}\) | A1 | Correct integration with correct limits seen (or implied for limits \(p\) and 10) and \(= \frac{1}{4}\) OE. Condone omission of \(\frac{3}{4000}\) |
| \(\frac{3}{4000}\left(1000 - \frac{1000}{3} - 100p + \frac{p^3}{3}\right) = \frac{1}{4}\) | M1 | Attempt substitution of correct limits in their integration of \(f(x)\). Accept limits 0 to \(p\) if clearly seen, accept limits \(-10\) and \(p\). Substitution must be seen. |
| \(\frac{2000}{3} - 100p + \frac{p^3}{3} = \frac{1000}{3}\), so \(p^3 - 300p + 1000 = 0\) | A1 | AG. No errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Curve is symmetrical about \(x = 0\) | B1 | May be implied by sketch. No contradictions or integrate \(f(x)\) between \(-q\) and \(+q\) and equate to 0.5 leading to \(q^3 - 300q + 1000 = 0\) oe |
| \(q = 3.47\) | B1 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{3}{4000}\int_5^{10}(100 - x^2)\,dx = \frac{3}{4000}\left[100x - \frac{x^3}{3}\right]_5^{10}$ | M1 | Attempt integration of $f(x)$, ignore limits. Condone omission of $\frac{3}{4000}$ |
| $= \frac{3}{4000}\left(1000 - \frac{1000}{3} - 500 + \frac{125}{3}\right)$ | M1 | Correct limits 5 and 10. OE SOI |
| $= 0.156$ (3 sf) or $\frac{5}{32}$ | A1 | For fully correct working seen including substitution of limits |
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{3}{4000}\int_p^{10}(100 - x^2)\,dx = \frac{1}{4}$ | M1 | Attempt integration of $f(x)$ with any limits and $= \frac{1}{4}$ or $= \frac{3}{4}$ seen. Condone omission of $\frac{3}{4000}$ |
| $\frac{3}{4000}\left[100x - \frac{x^3}{3}\right]_p^{10} = \frac{1}{4}$ | A1 | Correct integration with correct limits seen (or implied for limits $p$ and 10) and $= \frac{1}{4}$ OE. Condone omission of $\frac{3}{4000}$ |
| $\frac{3}{4000}\left(1000 - \frac{1000}{3} - 100p + \frac{p^3}{3}\right) = \frac{1}{4}$ | M1 | Attempt substitution of correct limits in their integration of $f(x)$. Accept limits 0 to $p$ if clearly seen, accept limits $-10$ and $p$. Substitution must be seen. |
| $\frac{2000}{3} - 100p + \frac{p^3}{3} = \frac{1000}{3}$, so $p^3 - 300p + 1000 = 0$ | A1 | AG. No errors seen |
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Curve is symmetrical about $x = 0$ | B1 | May be implied by sketch. No contradictions or integrate $f(x)$ between $-q$ and $+q$ and equate to 0.5 leading to $q^3 - 300q + 1000 = 0$ oe |
| $q = 3.47$ | B1 | |5 Bottles of Lanta contain approximately 300 ml of juice. The volume of juice, in millilitres, in a bottle is $300 + X$, where $X$ is a random variable with probability density function given by
$$f ( x ) = \begin{cases} \frac { 3 } { 4000 } \left( 100 - x ^ { 2 } \right) & - 10 \leqslant x \leqslant 10 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen bottle of Lanta contains more than 305 ml of juice.
\item Given that $25 \%$ of bottles of Lanta contain more than $( 300 + p ) \mathrm { ml }$ of juice, show that
$$p ^ { 3 } - 300 p + 1000 = 0 .$$
\item Given that $p = 3.47$, and that $50 \%$ of bottles of Lanta contain between ( $300 - q$ ) and ( $300 + q$ ) ml of juice, find $q$. Justify your answer.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q5 [9]}}