June 2019 Paper 3 Q2
2.
The partially completed box plot in Figure 1 shows the distribution of daily mean air temperatures using the data from the large data set for Beijing in 2015
An outlier is defined as a value
more than \(1.5 \times \text{IQR}\) below \(Q_1\) or
more than \(1.5 \times \text{IQR}\) above \(Q_3\)
The three lowest air temperatures in the data set are \(7.6\,{}^{\circ}\text{C}\), \(8.1\,{}^{\circ}\text{C}\) and \(9.1\,{}^{\circ}\text{C}\)
The highest air temperature in the data set is \(32.5\,{}^{\circ}\text{C}\)
Using the data from the large data set, Simon produced the following summary statistics for the daily mean air temperature, \(x\,{}^{\circ}\text{C}\), for Beijing in 2015
\[n = 184 \qquad \sum x = 4153.6 \qquad \mathrm{S}_{xx} = 4952.906\]Simon decides to model the air temperatures with the random variable
\[T \sim \mathrm{N}(22.6,\ 5.19^2)\]Simon wants to model another variable from the large data set for Beijing using a normal distribution.
| Scheme | Marks | AO |
|---|---|---|
| IQR \(= 26.6 - 19.4\ [= 7.2]\) | B1 | 2.1 |
| \(19.4 - 1.5 \times \text{‘}7.2\text{’}\ [= 8.6]\) or \(26.6 + 1.5 \times \text{‘}7.2\text{’}\ [= 37.4]\) | M1 | 1.1b |
| Plotting one upper whisker to 32.5 and one lower whisker to 8.6 or 9.1 | A1 | 1.1b |
| Plotting 7.6 and 8.1 as the only two outliers | A1 | 1.1b |
| (4) |
Notes
B1: for a correct calculation for the IQR (implied by 10.8 or 8.6 or 37.4 seen)
M1: for a complete method for either lower outlier limit or upper outlier limit (allow ft on their IQR) (may be implied by the 1st A1 or a lower whisker at 8.6)
A1: both whiskers plotted correctly (allow ½ square tolerance)
A1: only two outliers plotted, 7.6 and 8.1 (must be disconnected from whisker)
NOTE: A fully correct box plot with no incorrect working scores 4/4
| Scheme | Marks | AO |
|---|---|---|
| October (since it is the month with the coldest temperatures between May and October in Beijing) | B1 | 2.4 |
| (1) |
| Scheme | Marks | AO |
|---|---|---|
| \([\sigma =]\sqrt{\dfrac{4952.906}{184}}\) or e.g. \([\sigma =]\sqrt{\dfrac{\mathrm{S}_{xx}}{n}} = 5.188\ldots\) \([= 5.19^*]\) | B1cso* | 1.1b |
| (1) |
Notes
B1cso*: Correct expression with square root or correct formula and 5.188 or better
Allow a complete correct method finding \(\sum x^2 = \text{awrt } 98720\) and \(\sigma = \sqrt{\dfrac{98715.9\ldots}{184} - \left(\dfrac{4153.6}{184}\right)^2}\)
| Scheme | Marks | AO |
|---|---|---|
| \(z = (\pm)\ 1.28(16)\) or \([P_{90} =]\ 29.251\ldots\) or \([P_{10} =]\ 15.948\ldots\) | B1 | 3.1b |
| \(2 \times 1.2816 \times 5.19\) or ‘\(29.251\ldots\)’ – ‘\(15.948\ldots\)’ | M1 | 1.1b |
| \(=\) awrt 13.3 | A1 | 1.1b |
| (3) |
Notes
B1: Identifying \(z\)-value for 10th or 90th percentile (allow awrt \((\pm)\ 1.28\))
or for identifying \([P_{90} =]\ 29.251\ldots\) (awrt 29.3) or \([P_{10} =]\ 15.948\ldots\) (awrt 15.9)
(This may be implied by a correct answer awrt 13.3)
M1: for \(2 \times z \times 5.19\) where \(1 \lt z \lt 2\)
or for their \(P_{90} - P_{10}\) where \(25 \lt P_{90} \lt 35\) and \(10 \lt P_{10} \lt 20\)
A1: awrt 13.3
| Scheme | Marks | AO |
|---|---|---|
| Daily mean wind speed/Beaufort conversion since it is qualitative Rainfall since it is not symmetric/lots of days with 0 rainfall | B1 B1 | 2.4 2.4 |
| (2) | ||
| (11 marks) |
Notes
B1: for one variable identified and a correct supporting reason
B1: for two variables identified and a correct supporting reason for each
Allow any two of the following:
- Wind speed/Beaufort since the data is non-numeric (o.e.). They need not mention Beaufort provided there is a description of the data as non-numeric (Do not allow wind direction/wind gust)
- Rainfall as not symmetric/is skewed/is not bell shaped/lots of 0s /many days with no rain/mean≠mode or median
- Date since each data value appears once/it is uniformly distributed
- Daily mean pressure since it is not symmetric/is skewed/not bell shaped
- Daily mean wind speed since it is not symmetric/is skewed/not bell shaped
Ignore extraneous non-contradicting statements