| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Binomial |
| Difficulty | Standard +0.3 This is a standard chi-squared goodness of fit test with a binomial distribution. Students must calculate expected frequencies using B(5, 0.35), combine cells appropriately to meet the expected frequency ≥5 rule, compute the test statistic, and compare to critical value. While it requires multiple steps and careful arithmetic, it follows a well-rehearsed procedure with no novel insight required, making it slightly easier than average for Further Maths statistics. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 |
| Frequency | 12 | 39 | 46 | 37 | 12 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x\): 0, 1, 2, 3, 4, 5; Observed freq: 12, 39, 46, 37, 12, 4; Bin prob: 0.11603, 0.31236, 0.33642, 0.18115, 0.04877, 0.00525; Expected freq: 17.404, 46.858, 50.462, 27.172, 7.316, 0.7878 | M1 A1 | Attempt at E values (at least 4 correct); All correct to 2 dp or better |
| Add last two columns: 16, 8.104 | M1 | |
| \(\frac{(17.404-12)^2}{17.404} + \frac{(46.858-39)^2}{46.858} + \frac{(50.462-46)^2}{50.462} + \frac{(27.172-37)^2}{27.172} + \frac{(8.104-16)^2}{8.104} = 14.64\) or \(14.65\) | M1 A1 | Accept \(14.6 - 14.7\) |
| \(14.64 > 9.49\) | M1 | Compare their value with 9.49 |
| Freya's claim is not supported; data does not fit the distribution | A1 FT | Correct conclusion in context, FT only their 14.64; SC for Poisson M0M1M1M0 max 2 |
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x$: 0, 1, 2, 3, 4, 5; Observed freq: 12, 39, 46, 37, 12, 4; Bin prob: 0.11603, 0.31236, 0.33642, 0.18115, 0.04877, 0.00525; Expected freq: 17.404, 46.858, 50.462, 27.172, 7.316, 0.7878 | M1 A1 | Attempt at E values (at least 4 correct); All correct to 2 dp or better |
| Add last two columns: 16, 8.104 | M1 | |
| $\frac{(17.404-12)^2}{17.404} + \frac{(46.858-39)^2}{46.858} + \frac{(50.462-46)^2}{50.462} + \frac{(27.172-37)^2}{27.172} + \frac{(8.104-16)^2}{8.104} = 14.64$ or $14.65$ | M1 A1 | Accept $14.6 - 14.7$ |
| $14.64 > 9.49$ | M1 | Compare their value with 9.49 |
| Freya's claim is not supported; data does not fit the distribution | A1 FT | Correct conclusion in context, FT only their 14.64; SC for Poisson M0M1M1M0 max 2 |3 Apples are sold in bags of 5. Based on her previous experience, Freya claims that the probability of any apple weighing more than 100 grams is 0.35 , independently of other apples in the bag.
The apples in a random sample of 150 bags are checked and the number, $x$, in each bag weighing more than 100 grams is recorded. The results are shown in the following table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
Frequency & 12 & 39 & 46 & 37 & 12 & 4 \\
\hline
\end{tabular}
\end{center}
Carry out a goodness of fit test at the $5 \%$ significance level and hence comment on Freya's claim.\\
\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q3 [7]}}