| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2022 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Unbiased estimates then CI |
| Difficulty | Moderate -0.3 This is a straightforward confidence interval question requiring standard formulas for unbiased estimates and a normal distribution confidence interval with known σ. Part (c) adds mild interpretation but uses basic normal distribution properties. Slightly easier than average due to being a routine textbook exercise with clear steps. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = 11.42\) | B1 | for 11.42 cao |
| \(s^2 = \dfrac{1310.464 - 10 \times 11.42^2}{9}\) | M1 | for use of \(s^2 = \dfrac{\sum x^2 - n\bar{x}^2}{n-1}\) |
| \(= 0.7\) | A1 | for 0.7 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(z\) value for 95% CI is 1.96 | B1 | for writing or using 1.96 (or better from calculator 1.9599…) |
| \(11.42 \pm 1.96 \times \dfrac{0.8}{\sqrt{10}}\) | M1 | For use of \(\bar{x} \pm z \text{ value} \times \dfrac{\sigma}{\sqrt{n}}\) ft their \(z\) value, \(1 < |
| \((10.924\ldots,\ 11.915\ldots)\) awrt \((10.92, 11.92)\) | A1 A1 | for awrt 10.9 or awrt 11.9; for awrt 10.92 and awrt 11.92 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Y \sim N(``11.92\text{''}, 0.8^2)\) | M1 | for identifying the normal distribution with upper CI value as mean and 0.8 as standard deviation (may be seen in standardisation) |
| \(P(Y < 10.5) = P\!\left(Z < \dfrac{10.5 - ``11.92\text{''}}{0.8}\right) [= P(Z < -1.775)]\) | M1 | for standardising with 10.5, their mean (which must be in their CI including limits) from part (b) and standard deviation \(= 0.8\) |
| \(= 0.03837\ldots\) awrt 0.038 | A1 | awrt 0.038 (tables \(= 0.0375\)) |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = 11.42$ | B1 | for 11.42 cao |
| $s^2 = \dfrac{1310.464 - 10 \times 11.42^2}{9}$ | M1 | for use of $s^2 = \dfrac{\sum x^2 - n\bar{x}^2}{n-1}$ |
| $= 0.7$ | A1 | for 0.7 cao |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z$ value for 95% CI is 1.96 | B1 | for writing or using 1.96 (or better from calculator 1.9599…) |
| $11.42 \pm 1.96 \times \dfrac{0.8}{\sqrt{10}}$ | M1 | For use of $\bar{x} \pm z \text{ value} \times \dfrac{\sigma}{\sqrt{n}}$ ft their $z$ value, $1 < |z| < 2$ and their 11.42 |
| $(10.924\ldots,\ 11.915\ldots)$ awrt $(10.92, 11.92)$ | A1 A1 | for awrt 10.9 or awrt 11.9; for awrt 10.92 and awrt 11.92 |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Y \sim N(``11.92\text{''}, 0.8^2)$ | M1 | for identifying the normal distribution with upper CI value as mean and 0.8 as standard deviation (may be seen in standardisation) |
| $P(Y < 10.5) = P\!\left(Z < \dfrac{10.5 - ``11.92\text{''}}{0.8}\right) [= P(Z < -1.775)]$ | M1 | for standardising with 10.5, their mean (which must be in their CI including limits) from part (b) and standard deviation $= 0.8$ |
| $= 0.03837\ldots$ awrt 0.038 | A1 | awrt 0.038 (tables $= 0.0375$) |
---\begin{enumerate}
\item The weights, $x \mathrm {~kg}$, of each of 10 watermelons selected at random from Priya's shop were recorded. The results are summarised as follows
\end{enumerate}
$$\sum x = 114.2 \quad \sum x ^ { 2 } = 1310.464$$
(a) Calculate unbiased estimates of the mean and the variance of the weights of the watermelons in Priya's shop.
Priya researches the weight of watermelons, for the variety she has in her shop, and discovers that the weights of these watermelons are normally distributed with a standard deviation of 0.8 kg\\
(b) Calculate a $95 \%$ confidence interval for the mean weight of watermelons in Priya's shop. Give the limits of your confidence interval to 2 decimal places.
Priya claims that the confidence interval in part (b) suggests that nearly all of the watermelons in her shop weigh more than 10.5 kg\\
(c) Use your answer to part (b) to estimate the smallest proportion of watermelons in her shop that weigh less than 10.5 kg
\hfill \mbox{\textit{Edexcel S3 2022 Q1 [10]}}